mm-1008
===
Subject: Re: Please help. Having problem with simple math
logic
>I have 2 number scales.
>Top scale = 42 through -84
>and
>Bottom scale = 17 through 782
>If 17 on the bottom scale represents 42 on the top scale,
and the same
>goes for 782 on the bottom with -84 on the top, how do I
figure out
>any number on the top scale by inputting a number on the
bottom scale.
>(i.e. 261 on the bottom scale = ? on the top scale.
You mean e.g., not i.e. (Why do people use words when they
donÕt
know the meaning?)
But thereÕs a more substantive issue: You 
havenÕt told us what
relationship exists between the two scales. Is it supposed to
be a
straight-line relationship, one where e.g. any interval of 100
points on the bottom scale makes the same width interval on
the top
scale?
If you make that assumption, your problem can be solved. You
have
two (x,y) points, namely (17, 42) and (782, -48), and you
need to
find the equation of the line between them. The slope of that
line
is -90/765, so you use the point-slope form:
top = (-90/765)*(bottom-17) + 782
or
top = (-2/17)*bottom + 784
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Re: Please help. Having problem with simple math
logic
>>Top scale = 42 through -84
>>Bottom scale = 17 through 782
>If you make that assumption, your problem can be solved. You
have
>two (x,y) points, namely (17, 42) and (782, -48), and you
need to
>find the equation of the line between them. The slope of that
line
>is -90/765, so you use the point-slope form:
> top = (-90/765)*(bottom-17) + 782
Well _That_ was careless of me! The given point is (17,42),
so the
equation should be
top = (-90/765)(*bottom-17) + 42
or
top = (-2/17)*bottom + 44
which also works with the point (782, -48) as it should.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Re: ReL Please help. Having problem with simple
logic.
>>Top scale = 42 through -84
>>Bottom scale = 17 through 782
> The formula is T= (-0.1657)B+ 170.8. That a is negative is
due to
>the fact that the two scales are reversed, a higher value on
one
>corresponds to a lower value on the other.
The given point (17,42) does not work in that formula.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Re: Solving normal probability
>I need some serious help with probability. I just cannot get
it.
And what have you done to try to get it, other than simply
copy
the problem into your news posting software?
Please tell us what you tried, and _specifically_ where you 
got
stuck.
If you need a hint, hereÕs one: Your second and third
questions both
require you to apply the Empirical Rule.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Re: Yes, your field is corrupt
> Some of you may have noticed that IÕve finally 
focused on
the base
> assumption required to fight the proof that there is an over
one
> hundred year old definition in core mathematics, as posters
are
> finally revealed to be arguing for a relation like xy=uf,
where x and
> y are algebraic integer functions.
> ItÕs like someone arguing with you about xy=2, being in 
the
ring of
> integers, where x can be any integer, as I can even set
u=1, f=2, to
> *get* that value.
> So how have they gotten away with arguing such a position
for months?
> You and other mathematicians have let them because
mathematics is just
> a joke to you, and maybe a way to make money.
> Maybe you know that people just *trust* mathematicians, and
youÕre in
> the field to be in an area where the truth 
doesnÕt matter.
> Your field is corrupt. Mathematicians donÕt 
really care about
> anything but themselves. They are sick and lost.
> You are lost if you are with them.
> James Harris
> This is a crock of bull. If you are right, then I think
everyone who has
> taken math needs to go back and relearn the correct
mathematics. You need
to
> grow up; donÕt worry about this fame and money you claim
that you
deserve.
> You have a physics degree; USE IT!
> What would you do if someone attacked the field of physics?
ItÕd be the
same
> thing youÕre doing with mathematics. Get help......fast!
Oh yeah, try to shift the focus from your *proven* corruption.
ItÕs mathematicians who are defying mathematics.
ItÕs mathematicians who are now caught in a stupid lie to
hide an over
one hundred year old error.
ItÕs mathematicians who are running from the truth.
Your are corrupt and lost souls.
James Harris
===
Subject: Re: Implicit differentiation help
> Hi i also need help with an implicit differentiation
question. The
> question is to fin the 2nd derivative as a function of x if
sin y+cos
> y=x. I can get to a final answer but it i cant get rid of
the y
> values i have there. Could anyonehelp me get an answer in
terms of
I think this problem can be attacked primarily by going
through the
motions:
(1) sin(y) + cos(y) = x
Differentiate w.r.t. x:
(2) [cos(y) - sin(y)] * yÕ = 1
Multiply (2) by 1/yÕ to get:
(3) cos(y) - sin(y) = 1/yÕ
Now square (1) and (3) to get:
(4) sin^2(y) + 2*sin(y)*cos(y) + cos^2(y) = x^2
(5) sin^2(y) - 2*sin(y)*cos(y) + cos^2(y) = (1/yÕ)^2
The previous reorganization is a standard trick, so take
note. Now add (4)
and (5) to get:
(6) 2 = x^2 + (1/yÕ)^2
Isolate yÕ:
(7) yÕ = sqrt[1/(2 - x^2)]
ItÕs your job to finish. Note that the square 
root can be
positive or
negative.
There are other ways of arriving at this equation, but this
one is cleaner
that the first one I thought up. You should be able to proceed
from here.
Hope I didnÕt do too much of your homework for you!
.... Bob
===
Subject: Probability Question from a Computer Course
BACKGROUND:
part that modeled this situation (a sample output from that
part is at
the end of this post):
There are n=10 mines in the water.
A sonar can send one ping at a time, (which will reach all n
mines).
The ping will detect an individual mine with a probability
p=0.05.
Once a mine is detected, it stays detected forever (donÕt
need to
find it again, although a later ping may or may not detect 
it).
How long will it take to discover all 10 mines?
(The assignment had some other big parts to it; I didnÕt
write it to
find the repeated average of this part. I ran a few trials and
came up
with 67, 49, 46, 65, and 67 pings).
QUESTION:
How would one compute the actual expected value of the number
of pings
required? This was not a part of the assignment, but has been
bugging
me since IÕm also in a beginning probability course.
My thinking so far:
1. The expected number of detections in each ping is 1/2 a
mine, but
that doesnÕt seem to lead anywhere.
2. Each ping is a binomial process with n=10, p=0.05. So on
each ping,
I could calculate the probability of exactly r=0 detections,
r=1
detections, etc.
3. Now I have the probabilities of the 11 possible outcomes
(detected
r=0 mines, r=1, ... ,r=10) of the first ping from 2. I could
calculate
the conditional probabilities of the outcomes of the second
ping:
3.a. Given r=0 mines detected on the first ping, repeat 2.
3.b. Given r=1 mines detected on the first ping, calculate the
10
binomial probabilities of r=0..9 detections, with n=9, p=0.05.
3.c. Given r=2..r=10 on the first ping, similar to 3.b.
So, I would do something like 10! (or is it 10^10?)
computations of the
binomial formula, summing expected values along each branch
of a huge
tree. There must be a nicer way. Any suggestions?
*Sample Output of my program*
 
===
Subject: Re: Probability Question from a Computer Course
>BACKGROUND:
>part that modeled this situation (a sample output from that
part is at
>the end of this post):
>There are n=10 mines in the water.
>A sonar can send one ping at a time, (which will reach all n
mines).
>The ping will detect an individual mine with a probability
p=0.05.
>Once a mine is detected, it stays detected forever (donÕt
need to
>find it again, although a later ping may or may not detect
it).
>How long will it take to discover all 10 mines?
>(The assignment had some other big parts to it; I didnÕt
write it to
>find the repeated average of this part. I ran a few trials
and came up
>with 67, 49, 46, 65, and 67 pings).
>QUESTION:
>How would one compute the actual expected value of the
number of pings
>required? This was not a part of the assignment, but has
been bugging
>me since IÕm also in a beginning probability course.
YouÕll want to check all of the calculations, as I 
havenÕt had
time to do much double-checking.
For later convenience let q = 1 - p. Let p(k,r) be the
probability that exactly r mines remain undetected after the
first k pings. There are C(n,r) ways of choosing r mines. The
probability that a particular mine remains undetected after k
pings is q^k, so the probability that every mine in a
particular
set of r mines remains undetected after k pings is (q^k)^r =
q^(rk). The probability that a given mine is detected in at
most
k pings is 1 - q^k, so the probability that every mine in a
particular set of n - r mines is detected within k pings is
(1 - q^k)^(n-r). Thus,
p(k,r) = C(n,r)*q^(kr)*(1 - q^k)^(n-r).
In particular, p(k,0) = (1 - q^k)^n (which reassuringly does
approach 1 as k increases!). This isnÕt the probability that
it
will take k pings to detect all n mines, however, but rather
the
probability that it will take at most k pings. The
probability,
P(k), that it will take exactly k pings must be
P(k) = p(k,0) - p(k-1,0) = (1 - q^k)^n - (1 - q^(k-1))^n,
and e, the expected number of steps, must be
e = sum{k > 0; k[(1 - q^k)^n - (1 - q^(k-1))^n]}.
Consider the m-th partial sum e(m):
e(m) =
sum{k=1 to m; k[(1 - q^k)^n - (1 - q^(k-1))^n]} =
sum{k=1 to m; k(1 - q^k)^n} -
sum{k=1 to m; k(1 - q^(k-1))^n} =
sum{k=1 to m; k(1 - q^k)^n} -
sum{k=2 to m; k(1 - q^(k-1))^n} =
sum{k=1 to m; k(1 - q^k)^n} -
sum{k=1 to m-1; (k + 1)(1 - q^k)^n} =
sum{k=1 to m; k(1 - q^k)^n} -
sum{k=1 to m-1; k(1 - q^k)^n} -
sum{k=1 to m-1; (1 - q^k)^n} =
m(1 - q^m)^n - sum{k=1 to m-1; (1 - q^k)^n}.
Now
sum{k=1 to m-1; (1 - q^k)^n} =
sum{k=1 to m-1; sum{j; C(n,j) * (-1)^j * q^(jk)}} =
sum{j=0 to n; (-1)^j * C(n,j)*sum{k=1 to m-1; q^(jk)}} =
(m - 1) +
sum{j=1 to n; (-1)^j * C(n,j)*sum{k=1 to m-1; q^(jk)}} =
(m - 1) +
sum{j=0 to n; (-1)^j * C(n,j) * (q^j - q^jm)/(1 - q^j)}.
Similarly,
m(1 - q^m)^n =
m + sum{j=1 to n; m * C(n,j) * (-1)^j * q^(jm)},
so
e(m) =
1 + sum{j=1 to n; C(n,j)*(-1)^j*
[q^(jm)} - (q^j - q^(jm))/(1 - q^j]}.
Taking the limit as m increases without bound, we get
e = 1 - sum{j=1 to n; C(n,j) * (-1)^j * q^j / (1 - q^j)}. [*]
As a rough check, note that when n = 1 this reduces to
e = 1 + q/(1 - q) = 1/(1 - q) = 1/p,
as expected.
I donÕt immediately see a nice closed form for e in [*], but
this
at least reduces the problem considerably.
Brian
===
Subject: Re: addition
> I need help with adding two numbers (like 1, 2, and maybe
even
> others). HowÕs it possible to do it without a calculator?!
Also, what
> does it mean when the book says 1 + 1 + 1? IÕve only heard
about
> adding two numbers, not three. My calculus teacher wasnÕt
very clear
> about it.
It is considered an abomination to add a number other than 1
to another
number. So if you are confronted with say 50 + 20 you must
break one of
these
larger numbers down to a series of 1s to get back to basic
definitions.
(Hint,
choose the smaller one.) So we get 50 + (1 + 1 + 1 ..... +1)
You move the
first 1 outside the parentheses and you get 51 + (1+1+ ....
+1) You keep
going
like this one at a time and finally you will end up with 69 +
1 = 70. The
general result is known as the ones theorem. You must always
apply the ones
theroem whenever you have numbers to add together.
The reason your calculus teacher would not talk about adding
three numbers
is
that is also considered an abomination. This all goes back to
NoahÕs Ark
where
the animals were in twos - never threes. However, you can
achieve the same
result by applying the ones theorem in succession.
You should try to teach others about this, but you may have
better luck
outside of school where people have not been contaminated by
false ideas.
For
example, if you are at a truck stop on the way to University
and a waiter
is
adding more than two numbers and not applying the ones therom
you should
tell
him about the ones theorem and that what he is doing is an
abomination. Be
very persistant and I guarrantee you at the end there will be
a conversion.
Bill
===
Subject: Re: addition
>> I need help with adding two numbers (like 1, 2, and maybe
even
>> others). HowÕs it possible to do it without a 
calculator?!
Also, what
>> does it mean when the book says 1 + 1 + 1? IÕve only 
heard
about
>> adding two numbers, not three. My calculus teacher wasnÕt
very clear
>> about it.
> It is considered an abomination to add a number other than
1 to another
> number. So if you are confronted with say 50 + 20 you must
break one of
these
> larger numbers down to a series of 1s to get back to basic
definitions.
(Hint,
> choose the smaller one.) So we get 50 + (1 + 1 + 1 .....
+1) You move the
> first 1 outside the parentheses and you get 51 + (1+1+ ....
+1) You keep
going
> like this one at a time and finally you will end up with 69
+ 1 = 70. The
> general result is known as the ones theorem. You must
always apply the
ones
> theroem whenever you have numbers to add together.
Is this a consequence of the axioms that natural numbers,
have just read
them, or are you just messing with the OP?
> The reason your calculus teacher would not talk about
adding three numbers
is
> that is also considered an abomination. This all goes back
to NoahÕs Ark
where
> the animals were in twos - never threes. However, you can
achieve the
same
> result by applying the ones theorem in succession.
> You should try to teach others about this, but you may have
better luck
> outside of school where people have not been contaminated
by false ideas.
For
> example, if you are at a truck stop on the way to
University and a waiter
is
> adding more than two numbers and not applying the ones
therom you should
tell
> him about the ones theorem and that what he is doing is an
abomination.
Be
> very persistant and I guarrantee you at the end there will
be a
conversion.
> Bill
--
Sigblock empty. By choice.
===
Subject: Re: addition
> Is this a consequence of the axioms that natural numbers,
have just read
> them, or are you just messing with the OP?
Not all orÕs are exclusive.
Bill
===
Subject: Re: addition
>I donÕt know who youÕre trying to impress, 
but thereÕs no
way you can be
in
>calculus class and not know how to evaluate 1+1+1.
YouÕre probably right.
On the other hand, lots of my calculus (and statistics)
students
have been unable to evaluate sums like 37/1000 + 43/1000
correctly.
The answer 80/2000 is distressingly common.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: diagonalization
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9J2xcs05161;
The previous poster gave some historical remarks but did not
answer
the original posterÕs question. The diagonalization approach
will
fail for rationals because the new number created by altering
the
iÕth digit of each iÕth listed decimal will be 
irrational, not
rational. For proofs of the countability of the rationals,
see, for
instance, <a
href=http://www.amazon.com/exec/obidos/ASIN/0070381593/ref%
3Dnosim/fwsite1
-20/102-5225272-3993714>SchaumÕs
Outline of Set Theory</a>.
--g
===
Subject: getting rid of y
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9J2xcK05157;
One option would be to use a trig identity: Since you have
x = sin(y) + cos(y), squaring both sides yields:
x^2 = sin^2(y)+cos^2(y)+2sin(y)cos(y). Now, since
sin^2(y)+cos^2(y)=1, you have:
x^2-1 = 2sin(y)cos(y). By a trig identity, the right hand
side can be
re-written as sin(2y); so you have:
x^2-1=sin(2y), and now you can obtain y as an explicit
function of x
by taking inverse sin of both sides.
===
Subject: Re: getting rid of y
> One option would be to use a trig identity: Since you have
> x = sin(y) + cos(y), squaring both sides yields:
> x^2 = sin^2(y)+cos^2(y)+2sin(y)cos(y). Now, since
> sin^2(y)+cos^2(y)=1, you have:
> x^2-1 = 2sin(y)cos(y). By a trig identity, the right hand
side can be
> re-written as sin(2y); so you have:
> x^2-1=sin(2y), and now you can obtain y as an explicit
function of x
> by taking inverse sin of both sides.
Slick.
===
Subject: Help on a proof for convergence?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9J2xb805149;
How would one go about proving the convergence of the
following
series:
sin(n)/n
? There is no interval given. Most of the tests used for
proving
convergence cannot be used: alternating series (because the
terms are
not all decreasing, as is the case if it was only 1 to
infinity), the
comparison tests (you canÕt compare the series to 1/n), and
the root
and ratio tests simply make the series more difficult. So does
anyone
know how to prove this? Any help would be appreciated.
===
Subject: Re: Help on a proof for convergence?
> How would one go about proving the convergence of the
following
> series:
> sin(n)/n
> ? There is no interval given. Most of the tests used for
proving
> convergence cannot be used: alternating series (because the
terms are
> not all decreasing, as is the case if it was only 1 to
infinity), the
> comparison tests (you canÕt compare the series to 1/n), 
and
the root
> and ratio tests simply make the series more difficult. So
does anyone
> know how to prove this? Any help would be appreciated.
Hi Johnny, nice question !
I understand that you need the convergence of
S:= SUM_{k=1 to k= infty}sin(k)/k .
Try Fourier series of certain odd function f(x) such that
f(x)= SUM_{k=1 to k=infty} sin(kx)/k . Then , if possible put
x=1.
Perhaps help. Alex
Note that S_n(x):=SUM_{k=1 to k=n} sin(kx)/k >0 for all x in
(0,pi) .
This is FejerÕs inequality (1905 ?) .
I think that you must take the 2*pi periodic odd function
f:R-->R
whose restriction at (-pi,pi] is defined as
f(x)= (x-pi)/pi for x in (-pi ,0)
f(x)= 0 when x=0
f(x)= (pi-x)/pi for x in (0,pi] .
If you look in a treatise on trigonometric series (A.Zygmund
or ?.Edwards)
you will see that the convergence is assured. Therefore, my
impresion
is that the sum of your series is S=f(1)=(pi-1)/pi = 1- 1/pi .
===
Subject: Re: Help on a proof for convergence?
Not helpful on any proofs, but you may find this link
interesting...
http://mathworld.wolfram.com/SincFunction.html
--
Dana
= = = = = = = = = = = = = = = = =
> How would one go about proving the convergence of the
following
> series:
> sin(n)/n
> ? There is no interval given. Most of the tests used for
proving
> convergence cannot be used: alternating series (because the
terms are
> not all decreasing, as is the case if it was only 1 to
infinity), the
> comparison tests (you canÕt compare the series to 1/n), 
and
the root
> and ratio tests simply make the series more difficult. So
does anyone
> know how to prove this? Any help would be appreciated.
===
Subject: Re: Help on a proof for convergence?
>How would one go about proving the convergence of the
following
>series:
>sin(n)/n
>? There is no interval given. Most of the tests used for
proving
>convergence cannot be used: alternating series (because the
terms are
>not all decreasing, as is the case if it was only 1 to
infinity), the
>comparison tests (you canÕt compare the series to 1/n),
CanÕt you???
Obviously the denominators are equal. CanÕt you compare the
numerators?
-1 < sin(n) < 1 for any integral n. N.B. <, not <=. You know
this
because the only time sin(x) = +1 or -1 is when x = k(pi/2)+1
for
integer k. Since pi itself is irrational, k(pi/2)+1 can never
be an
integer; therefore sin(n) for integer n is never equal to +1
or -1.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Re: Help on a proof for convergence?
>>How would one go about proving the convergence of the
following
>>series:
Sorry, I misread that as _sequence_. Please disregard my
previous
follow-up.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Re: Help on a proof for convergence?
> How would one go about proving the convergence of the
following
> series:
> sin(n)/n
> ? There is no interval given.
I donÕt understand. Do you mean that the initial and 
final
values of the
index n in the summation where not specified? If so, then
perhaps you
were to assume that n goes from 1 to infinity. (If 
thatÕs not
true, then
more information is needed.)
> Most of the tests used for proving convergence cannot be
used:
> alternating series (because the terms are not all
decreasing, as is the
> case if it was only 1 to infinity),
Right (despite the fact that MadJock seems to think that the
terms
decrease).
> the comparison tests (you canÕt compare the series to 
1/n),
and the root
> and ratio tests simply make the series more difficult. So
does anyone
> know how to prove this? Any help would be appreciated.
If n goes from 1 to infinity, the series converges to (pi -
1)/2.
That series arises fairly frequently in math newsgroups. Do a
Google Groups
search for
sin(n)/n series convergence
Of course youÕll get some stuff which isnÕt 
pertinent, but
youÕll also
get what you need. The second item I found was an excellent
response by
Rob Johnson, for example.
BTW, if perchance the summation was instead supposed to be
over _all_
integers n (from -infinity to +infinity), then you 
would most
likely be
supposed to take sin(n)/n to be 1 when n = 0, and the sum
would then be,
neatly, just pi.
David Cantrell
===
Subject: Re: Help on a proof for convergence?
===
Subject: Help on a proof for convergence?
>How would one go about proving the convergence of the
following
>series: sin(n)/n
lim(n->oo) (sin n)/n = 0
Proof:
For eps > 0, find n0 with 1/n0 < eps
Thus for all n > n0,
|(sin n)/n - 0 | = |(sin n)/n| <= 1/n < 1/n0 < eps
----
===
Subject: Re: Help on a proof for convergence?
===
> Subject: Help on a proof for convergence?
> >How would one go about proving the convergence of the
following
> >series: sin(n)/n
ItÕs obvious that the _sequence_ sin(n)/n converges. But
notice the word
series above, William. Establishing convergence of the series
is not so
simple.
David
> lim(n->oo) (sin n)/n = 0
> Proof:
> For eps > 0, find n0 with 1/n0 < eps
> Thus for all n > n0,
> |(sin n)/n - 0 | = |(sin n)/n| <= 1/n < 1/n0 < eps
===
Subject: Re: Help on a proof for convergence?
> >How would one go about proving the convergence of the
following
> >series: sin(n)/n
> ItÕs obvious that the _sequence_ sin(n)/n converges. But
notice the word
> series above, William. Establishing convergence of the
series is not
so
> simple.
Oh, OP wants to see the series
sum_n (sin n)/n
converge. Well now that thatÕs unambigious.
N modulus 2pi, is dense in the reals [0,2pi).
Thus on the average itÕs an alternating series
which makes it convergent, and also complicated.
===
Subject: Re: Help on a proof for convergence?
Weird. Seems quite obvious in my head that it converges too -
itÕs a
common
function in digital transmission systems. As n increases,
f(n) decreases.
But I donÕt know how to prove it. Sorry.
MadJock
> How would one go about proving the convergence of the
following
> series:
> sin(n)/n
> ? There is no interval given. Most of the tests used for
proving
> convergence cannot be used: alternating series (because the
terms are
> not all decreasing, as is the case if it was only 1 to
infinity), the
> comparison tests (you canÕt compare the series to 1/n), 
and
the root
> and ratio tests simply make the series more difficult. So
does anyone
> know how to prove this? Any help would be appreciated.
===
Subject: how would 4 9Õs equal 100 using basic math
calculations?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9J62wC16584;
I would like to ask for your help in solving a mathematical
problem
that has me stumped. Lets say you have four 9Õs. Now, using
any basic
arithmetical procedure (multiply, add, divide, subtract), how
would
you get them to equal 100. Is this possible or would this be
a trick
question?
Any help would be greatly appreciated.
Eric
===
Subject: Update to problem, my actual work done
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9JBpEV07258;
I would like to apologize to everybody for not posting the
work I
actually did to attempt to solve this problem. Here is what I
have
done so far. Unless this is a trick question, or my math is
poor, does
this have a solution?
Addition
9 + 9 = 18 + 9 = 27 + 9 = 36
9 + 9 = 18 + 9 = 27 - 9 = 18
9 + 9 = 18 + 9 = 27 * 9 = 243
9 + 9 = 18 + 9 = 27 / 9 = 3
9 + 9 = 18 - 9 = 9 + 9 = 18
9 + 9 = 18 - 9 = 9 - 9 = 0
9 + 9 = 18 - 9 = 9 * 9 = 81
9 + 9 = 18 - 9 = 9 / 9 = 1
9 + 9 = 18 * 9 = 169 + 9 = 176
9 + 9 = 18 * 9 = 169 - 9 = 160
9 + 9 = 18 * 9 = 169 * 9 = 1521
9 + 9 = 18 * 9 = 169 / 9 = 18.77777
9 + 9 = 18 / 9 = 9 + 9 = 18
9 + 9 = 18 / 9 = 9 - 9 = 0
9 + 9 = 18 / 9 = 9 * 9 = 81
9 + 9 = 18 / 9 = 9 / 9 = 1
Subtraction
9 - 9 = 0 + 9 = 9 + 9 = 18
9 - 9 = 0 + 9 = 9 - 9 = 0
9 - 9 = 0 + 9 = 9 * 9 = 81
9 - 9 = 0 + 9 = 9 / 9 = 1
9 - 9 = 0 - 9 = -9 + 9 = 0
9 - 9 = 0 - 9 = -9 - 9 = -18
9 - 9 = 0 - 9 = -9 * 9 = -81
9 - 9 = 0 - 9 = -9 / 9 = -1
9 - 9 = 0 * 9 = 0 + 9 = 9
9 - 9 = 0 * 9 = 0 - 9 = -9
9 - 9 = 0 * 9 = 0 * 9 = 0
9 - 9 = 0 * 9 = 0 / 9 = 0
9 - 9 = 0 / 9 = 0 + 9 = 9
9 - 9 = 0 / 9 = 0 - 9 = -9
9 - 9 = 0 / 9 = 0 * 9 = 81
9 - 9 = 0 / 9 = 0 / 9 = 9
Multiplication
9 * 9 = 81 + 9 = 90 + 9 = 99
9 * 9 = 81 + 9 = 90 - 9 = 81
9 * 9 = 81 + 9 = 90 * 9 = 810
9 * 9 = 81 + 9 = 90 / 9 = 10
9 * 9 = 81 - 9 = 72 + 9 = 81
9 * 9 = 81 - 9 = 72 - 9 = 63
9 * 9 = 81 - 9 = 72 * 9 = 648
9 * 9 = 81 - 9 = 72 / 9 = 8
9 * 9 = 81 * 9 = 729 + 9 = 738
9 * 9 = 81 * 9 = 729 - 9 = 720
9 * 9 = 81 * 9 = 729 * 9 = 6561
9 * 9 = 81 * 9 = 729 / 9 = 81
9 * 9 = 81 / 9 = 9 + 9 = 18
9 * 9 = 81 / 9 = 9 - 9 = 0
9 * 9 = 81 / 9 = 9 * 9 = 81
9 * 9 = 81 / 9 = 9 / 9 = 1
Division
9 / 9 = 1 + 9 = 10 + 9 = 19
9 / 9 = 1 + 9 = 10 - 9 = 1
9 / 9 = 1 + 9 = 10 * 9 = 90
9 / 9 = 1 + 9 = 10 / 9 = 1.1111
9 / 9 = 1 - 9 = -8 + 9 = 1
9 / 9 = 1 - 9 = -8 - 9 = -17
9 / 9 = 1 - 9 = -8 * 9 = -72
9 / 9 = 1 - 9 = -8 / 9 = -0.8888
9 / 9 = 1 * 9 = 9 + 9 = 18
9 / 9 = 1 * 9 = 9 - 9 = 0
9 / 9 = 1 * 9 = 9 * 9 = 81
9 / 9 = 1 * 9 = 9 / 9 = 1
9 / 9 = 1 / 9 = 0.1111 + 9 = 9.1111
9 / 9 = 1 / 9 = 0.1111 - 9 = -8.8888
9 / 9 = 1 / 9 = 0.1111 * 9 = 1
9 / 9 = 1 / 9 = 0.1111 / 9 = 0.0123456790
Eric
===
Subject: Re: how would 4 9Õs equal 100 using basic math
calculations?
> I would like to ask for your help in solving a mathematical
problem
> that has me stumped. Lets say you have four 9Õs. Now, 
using
any basic
> arithmetical procedure (multiply, add, divide, subtract),
how would
> you get them to equal 100. Is this possible or would this
be a trick
> question?
99 + 9/9
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9JBpDM07210;
- when you square both sides of the original equation how do
you end
up with 2sin(y)cos(y) as I cannot see where that has come
from -
probably due to ignorance or the fact i have overlooked
something. One
other note just to be sure yogi how did the squareÕs
disappear from
sin(y)yÕ^2 and cos(y)yÕ^2 when u took them 
across to the
other side -
i just need these points to check that i can go through the
process
>- when you square both sides of the original equation how do
you end
>up with 2sin(y)cos(y) as I cannot see where that has come
from -
Neither can I, because of the way you posted. Much as IÕd
like to
help, I canÕt.
When youÕre posting a follow-up, post it _as_ a follow-up.
Your
youÕre referring to.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Dont worry about the finding 2sinycosy bit
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9JBpDZ07218;
I was just being stupid - forgot for a moment that the whole
thing was
squared rather than the individual bits
===
Subject: Re: Quick Math Guide to core error issues
James,
You post so much material that I donÕt have the time or
desire to
read everything and I lose track of your position and
arguments.
Some time ago you had an organised website with definitions
and cross-
reference and you have posted several summaries and histories
on this
newsgroup which would be easier to refer to if they were
permanently
available. I, for one, would be pleased if you would
resurrect your
website and put the information back.
Here are just some suggestions as to what you could put there
and
how it could be organised.
*Definitions*
- Algebraic Integers and Algebraic Numbers
An explanation of what an algebraic integer is , some
examples and their general properties would be useful.
I vaguely recall that you had some software that would
take a monic quadratic equation with integer coefficients
and factorise it into linear terms with algebraic integer
coefficients. This could usefully go here as a Java applet.
- Your ŌObject RingÕ and what numbers are in it 
and what
arenÕt.
I *think* you said 2^sqrt(3) was in it
I *think* you said that it wasnÕt wholly contained in the
complex numbers
An explanation of why these numbers are in it would be useful
- What an incomplete ring is.
- Some expressions donÕt seem to translate across the
Atlantic and
I am thinking of Ōhas a factor of 5Õ in 
particular. To me,
Ōhas a
value of 42Õ means Ō42 is a 
valueÕ so Ōhas a factor of 5Õ
should
mean Ō5 is a factorÕ but you 
donÕt seem to mean this all time.
Personally, I prefer the active Ō5 divides NÕ or 
perhaps ŌN is
divisible by 5Õ to the passive ŌN has 5 as a 
factorÕ.
*History*
I think you have written at least two histories to this
newsgroup
and a web site would be a much better place to put them. You
have
made several accusations of lying and you could use the
website
to substantiate them by pointing to documentation.
*Points of Contention*
Your Viewpoint
I lose track of the arguments and how they are settled or even
if they are. Your executive summary below is a good start but
that
is all it is. Hotlinks to expansions, with proofs, would be
better.
Java documentation is a fairly good example if the analogy
isnÕt
taken too far - at the top of the documentation for a class is
an explanation of what this class is for and a list of the
methods
with the types of their parameters and each has a hotlink to
an
explanation of that method. One could go further and look at
the
code but it should never be necessary.
Similarly, a mathematical exposition can have definitions,
lemmas,
propositions and theorems. The proof of a lemma or similar
can cite
consequences from the statement of another but mustnÕt ever
refer to
the proof - this would be as bad as code jumping into the
middle of
a subroutine or the documentation of a method referring to the
code of another method.
I suggest an exposition of your work would have the same
structure
with an executive summary at the top and cascading expansions
of
each point. (A good book contains a series of chapters. Each
chapter
starts out by saying what it is going to say, then says it in
detail,
and finally ends by summarising what it has said) Your 
previous
website
made a point of being terse but this isnÕt necessary.
Other Viewpoints
It is not at all an easy thing to do, but if you could try to
explain,
in detail, other peopleÕs standpoints as well as your own
then this
would be useful. IÕm thinking in particular of the
factorisation
of certain polynomials in which you say that one root is
coprime
to a prime factor of the constant term of the polynomial and
other
people have shown explicit factorisations of the polynomial in
which this is not true. I didnÕt follow your rebuttal.
Dead Viewpoints
It would take great courage but if you could also document
points
where you are no longer holding a former position then this
also
would interesting. IÕm thinking in particular of whether a
ring
must be closed under an infinite summation and whether Z[pi] 
is
isomorphic to the reals.
*Simplifications*
The polynomial that you use is very complicated with several
parameters. Is all the complication necessary to the proof
and does
it add to it? Is the ŌuÕ term necessary at all? 
You have
resorted
to actually substituting values in order to explain - why not
go
back and give the whole argument again but use a polynomial
which
is as simple as possible. In fact, I donÕt always follow 
your
reasoning but perhaps it would be more obvious if you used a
quadratic polynomial instead of a trinomial or explained why
the
argument doesnÕt work for quadratics.
*Style*
I, too, like long sentences (my second choice for a book to
take
if I were to be stranded on a desert island would be the
collected
works of Jane Austen) and you are much better at writing them
than
I am but they can be difficult for other people to follow.
Take this as an example:
. I already have as the polynomial is
.
. P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3f)
.
. and your attempts at confusing the issue by trying to push
up x,
. donÕt help you here, as the gÕs as factors 
of m,
necessarily have
. a constant term with respect to m, and your claim that it
can vary
. with m, is nonsensical on its face.
It makes sense and I can understand it but I have to work at
it.
I suggest that shorter sentences would make it easier for the
reader.
Penny Hassett
> For those of you trying to keep up with the mathematical
facts in the
> discussions about the error in core mathematics from a
problem with a
> definition, this post will outline the important ones
quickly and
> succinctly.
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
> 2. The important tool I use is a polynomial:
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3 f)
> The form of the polynomial allows me to factor P(m) into
> non-polynomial factors, and the factorization with those
factors is
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> where the aÕs are roots of the following cubic:
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
> 3. Dispute centers around what happens when I divide P(m)
by f^2,
> which youÕll note is a factor of the polynomial in the 
ring
of
> algebraic integers.
> 4. Mathematicians have argued that f^2 divides off as a
function of m
> because if they concede that it divides off independent of
m, then I
> can show that only two of the roots of
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)
> have f as a factor.
> 5. However, it turns out that if you go to the field of
algebraic
> numbers you can prove that for *certain* values of m and f,
the roots
> of the cubic do not have f as a factor *in the ring of
algebraic
> numbers* which is the inconsistency.
> That is, for the math to be consistent, two of the roots
*should* have
> f as a factor as long as m and f are algebraic integers,
but while I
> can show they do for a particular values like m=1,
f=sqrt(2), there
> are other values you can show they do not *in the ring of
algebraic
> integers* which results from the definition and its focus on
monic
> polynomials.
> Note: In the ring of algebraic integers you canÕt see the
problem but
> have to go to the field of algebraic numbers as from within
the ring
> of algebraic integers it appears that only two of the roots
have a
> factor that is f.
> James Harris
===
Subject: Re: Quick Math Guide to core error issues
> James,
> You post so much material that I donÕt have the time or
desire to
> read everything and I lose track of your position and
arguments.
> Some time ago you had an organised website with definitions
and cross-
> reference and you have posted several summaries and
histories on this
> newsgroup which would be easier to refer to if they were
permanently
> available. I, for one, would be pleased if you would
resurrect your
> website and put the information back.
> Here are just some suggestions as to what you could put
there and
> how it could be organised.
> *Definitions*
> - Algebraic Integers and Algebraic Numbers
> An explanation of what an algebraic integer is , some
> examples and their general properties would be useful.
> I vaguely recall that you had some software that would
> take a monic quadratic equation with integer coefficients
> and factorise it into linear terms with algebraic integer
> coefficients. This could usefully go here as a Java applet.
> - Your ŌObject RingÕ and what numbers are in 
it and what
arenÕt.
> I *think* you said 2^sqrt(3) was in it
> I *think* you said that it wasnÕt wholly contained in the
> complex numbers
> An explanation of why these numbers are in it would be
useful
> - What an incomplete ring is.
> - Some expressions donÕt seem to translate across the
Atlantic and
> I am thinking of Ōhas a factor of 5Õ in 
particular. To me,
Ōhas a
> value of 42Õ means Ō42 is a 
valueÕ so Ōhas a factor of 5Õ
should
> mean Ō5 is a factorÕ but you 
donÕt seem to mean this all
time.
> Personally, I prefer the active Ō5 divides NÕ 
or perhaps ŌN
is
> divisible by 5Õ to the passive ŌN has 5 as a 
factorÕ.
> *History*
> I think you have written at least two histories to this
newsgroup
> and a web site would be a much better place to put them.
You have
> made several accusations of lying and you could use the
website
> to substantiate them by pointing to documentation.
> *Points of Contention*
> Your Viewpoint
> I lose track of the arguments and how they are settled or
even
> if they are. Your executive summary below is a good start
but that
> is all it is. Hotlinks to expansions, with proofs, would be
better.
> Java documentation is a fairly good example if the analogy
isnÕt
> taken too far - at the top of the documentation for a class
is
> an explanation of what this class is for and a list of the
methods
> with the types of their parameters and each has a hotlink
to an
> explanation of that method. One could go further and look
at the
> code but it should never be necessary.
> Similarly, a mathematical exposition can have definitions,
lemmas,
> propositions and theorems. The proof of a lemma or similar
can cite
> consequences from the statement of another but mustnÕt 
ever
refer to
> the proof - this would be as bad as code jumping into the
middle of
> a subroutine or the documentation of a method referring to
the
> code of another method.
> I suggest an exposition of your work would have the same
structure
> with an executive summary at the top and cascading
expansions of
> each point. (A good book contains a series of chapters.
Each chapter
> starts out by saying what it is going to say, then says it
in detail,
> and finally ends by summarising what it has said) Your
previous
> website
> made a point of being terse but this isnÕt necessary.
> Other Viewpoints
> It is not at all an easy thing to do, but if you could try
to
> explain,
> in detail, other peopleÕs standpoints as well as your own
then this
> would be useful. IÕm thinking in particular of the
factorisation
> of certain polynomials in which you say that one root is
coprime
> to a prime factor of the constant term of the polynomial
and other
> people have shown explicit factorisations of the polynomial
in
> which this is not true. I didnÕt follow your rebuttal.
> Dead Viewpoints
> It would take great courage but if you could also document
points
> where you are no longer holding a former position then this
also
> would interesting. IÕm thinking in particular of whether a
ring
> must be closed under an infinite summation and whether Z[pi]
is
> isomorphic to the reals.
>
> *Simplifications*
> The polynomial that you use is very complicated with several
> parameters. Is all the complication necessary to the proof
and does
> it add to it? Is the ŌuÕ term necessary at 
all? You have
resorted
> to actually substituting values in order to explain - why
not go
> back and give the whole argument again but use a polynomial
which
> is as simple as possible. In fact, I donÕt always follow
your
> reasoning but perhaps it would be more obvious if you used a
> quadratic polynomial instead of a trinomial or explained
why the
> argument doesnÕt work for quadratics.
> *Style*
>
> I, too, like long sentences (my second choice for a book to
take
> if I were to be stranded on a desert island would be the
collected
> works of Jane Austen) and you are much better at writing
them than
> I am but they can be difficult for other people to follow.
> Take this as an example:
> . I already have as the polynomial is
> . P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x
u^2 + u^3f)
> . and your attempts at confusing the issue by trying to
push up x,
> . donÕt help you here, as the gÕs as factors 
of m,
necessarily have
> . a constant term with respect to m, and your claim that it
can vary
> . with m, is nonsensical on its face.
> It makes sense and I can understand it but I have to work
at it.
> I suggest that shorter sentences would make it easier for
the
> reader.
> Penny Hassett
>
> For those of you trying to keep up with the mathematical
facts in the
> discussions about the error in core mathematics from a
problem with a
> definition, this post will outline the important ones
quickly and
> succinctly.
>
> 1. First the problematic definition:
>
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
>
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
>
> 2. The important tool I use is a polynomial:
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3 f)
>
> The form of the polynomial allows me to factor P(m) into
> non-polynomial factors, and the factorization with those
factors is
>
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> where the aÕs are roots of the following cubic:
>
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
>
> 3. Dispute centers around what happens when I divide P(m)
by f^2,
> which youÕll note is a factor of the polynomial in the 
ring
of
> algebraic integers.
>
> 4. Mathematicians have argued that f^2 divides off as a
function of m
> because if they concede that it divides off independent of
m, then I
> can show that only two of the roots of
>
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)
>
> have f as a factor.
>
> 5. However, it turns out that if you go to the field of
algebraic
> numbers you can prove that for *certain* values of m and f,
the roots
> of the cubic do not have f as a factor *in the ring of
algebraic
> numbers* which is the inconsistency.
>
> That is, for the math to be consistent, two of the roots
*should* have
> f as a factor as long as m and f are algebraic integers,
but while I
> can show they do for a particular values like m=1,
f=sqrt(2), there
> are other values you can show they do not *in the ring of
algebraic
> integers* which results from the definition and its focus on
monic
> polynomials.
>
> Note: In the ring of algebraic integers you canÕt see the
problem but
> have to go to the field of algebraic numbers as from within
the ring
> of algebraic integers it appears that only two of the roots
have a
> factor that is f.
>
> James Harris
What a breath of fresh air you are on this NG!
Respectfully,
John
===
Subject: Re: Quick Math Guide to core error issues
> James,
> You post so much material that I donÕt have the time or
desire to
> read everything and I lose track of your position and
arguments.
> Some time ago you had an organised website with definitions
and cross-
> reference and you have posted several summaries and
histories on this
> newsgroup which would be easier to refer to if they were
permanently
> available. I, for one, would be pleased if you would
resurrect your
> website and put the information back.
I like you Penny Hasset and appreciate your commentary which
is why
IÕm posting in a thread where I donÕt need to 
post as itÕs a
quide.
My problem is that I donÕt want to use MSN Groups, and I
donÕt feel
like going to another website provider.
I *am* willing to allow someone else to host my work as long
as they
give me complete editorial control.
Otherwise, itÕs easier for me to just post rather than try 
to
maintain
a website.
> Here are just some suggestions as to what you could put
there and
> how it could be organised.
> *Definitions*
> - Algebraic Integers and Algebraic Numbers
> An explanation of what an algebraic integer is , some
> examples and their general properties would be useful.
> I vaguely recall that you had some software that would
> take a monic quadratic equation with integer coefficients
> and factorise it into linear terms with algebraic integer
> coefficients. This could usefully go here as a Java applet.
Oh yeah, after Arturo Magidin tried to make a big deal out of
some
crap, I figured out what he was doing, which 
wasnÕt much more
than a
rather simple search for a factorization in algebraic
integers.
It was fun, but not that much fun.
You see, IÕve found and dropped more mathematics than most
people
discover in a lifetime, as IÕm a thrill seeker.
You know, an adrenaline junkie.
> - Your ŌObject RingÕ and what numbers are in 
it and what
arenÕt.
> I *think* you said 2^sqrt(3) was in it
> I *think* you said that it wasnÕt wholly contained in the
> complex numbers
> An explanation of why these numbers are in it would be
useful
Actually *you* are in it Penny Hasset, as the object ring is
rather
large.
You see, you are a mathematical object which I can prove
using some
rather basic logic and GoedelÕs proof.
I like it that youÕre in Britain.
If youÕre willing to advise me, IÕm willing to 
toe a line.
After all, it is math, but by myself I tend to be over the
top.
I need organization.
> - What an incomplete ring is.
ItÕs a ring where you can have contradictions *within* the
ring, which
is the problem with algebraic integers.
I can explain everything, but IÕm a discoverer. 
IÕm an artist.
IÕm NOT organized for this other stuff, like trying to
convince
people.
IÕm an artist.
> - Some expressions donÕt seem to translate across the
Atlantic and
> I am thinking of Ōhas a factor of 5Õ in 
particular. To me,
Ōhas a
> value of 42Õ means Ō42 is a 
valueÕ so Ōhas a factor of 5Õ
should
> mean Ō5 is a factorÕ but you 
donÕt seem to mean this all
time.
> Personally, I prefer the active Ō5 divides NÕ 
or perhaps ŌN
is
> divisible by 5Õ to the passive ŌN has 5 as a 
factorÕ.
Hey Penny Hasset, if you can help me, and help me overcome
these
objections to the extent that I can make some money here,
IÕll pay you
$250,000 US from any one math prize that I win that exceeds
that
amount.
Since IÕm a black male in America, as it has a rather
stupendous
history of racism, I should be able to pay that amount as
well as
$100,000 US as previously offered to a person or group with a
machine
proof of the core error, without much trouble, since I turn
the world
upside down.
> *History*
> I think you have written at least two histories to this
newsgroup
> and a web site would be a much better place to put them.
You have
> made several accusations of lying and you could use the
website
> to substantiate them by pointing to documentation.
Oh, thatÕs part of my fun. Unfortunately for me I have a
tendency to
scare people away when they figure out just how much I know
and what I
can do.
Luckily for me, mathematicians are arrogant *and* dumb.
So theyÕre a perfect combination for someone like me, who
otherwise
gets kind of lonely.
> *Points of Contention*
> Your Viewpoint
> I lose track of the arguments and how they are settled or
even
> if they are. Your executive summary below is a good start
but that
> is all it is. Hotlinks to expansions, with proofs, would be
better.
> Java documentation is a fairly good example if the analogy
isnÕt
> taken too far - at the top of the documentation for a class
is
> an explanation of what this class is for and a list of the
methods
> with the types of their parameters and each has a hotlink
to an
> explanation of that method. One could go further and look
at the
> code but it should never be necessary.
> Similarly, a mathematical exposition can have definitions,
lemmas,
> propositions and theorems. The proof of a lemma or similar
can cite
> consequences from the statement of another but mustnÕt 
ever
refer to
> the proof - this would be as bad as code jumping into the
middle of
> a subroutine or the documentation of a method referring to
the
> code of another method.
> I suggest an exposition of your work would have the same
structure
> with an executive summary at the top and cascading
expansions of
> each point. (A good book contains a series of chapters.
Each chapter
> starts out by saying what it is going to say, then says it
in detail,
> and finally ends by summarising what it has said) Your
previous
> website
> made a point of being terse but this isnÕt necessary.
I agree. LetÕs get started.
I can make you rich, if you arenÕt rich already.
If you are rich, IÕll make you powerful.
If youÕre already powerful, hell, why not just do it?
> Other Viewpoints
> It is not at all an easy thing to do, but if you could try
to
> explain,
> in detail, other peopleÕs standpoints as well as your own
then this
> would be useful. IÕm thinking in particular of the
factorisation
> of certain polynomials in which you say that one root is
coprime
> to a prime factor of the constant term of the polynomial
and other
> people have shown explicit factorisations of the polynomial
in
> which this is not true. I didnÕt follow your rebuttal.
> Dead Viewpoints
> It would take great courage but if you could also document
points
> where you are no longer holding a former position then this
also
> would interesting. IÕm thinking in particular of whether a
ring
> must be closed under an infinite summation and whether Z[pi]
is
> isomorphic to the reals.
>
> *Simplifications*
> The polynomial that you use is very complicated with several
> parameters. Is all the complication necessary to the proof
and does
> it add to it? Is the ŌuÕ term necessary at 
all? You have
resorted
> to actually substituting values in order to explain - why
not go
> back and give the whole argument again but use a polynomial
which
> is as simple as possible. In fact, I donÕt always follow
your
> reasoning but perhaps it would be more obvious if you used a
> quadratic polynomial instead of a trinomial or explained
why the
> argument doesnÕt work for quadratics.
> *Style*
>
> I, too, like long sentences (my second choice for a book to
take
> if I were to be stranded on a desert island would be the
collected
> works of Jane Austen) and you are much better at writing
them than
> I am but they can be difficult for other people to follow.
> Take this as an example:
> . I already have as the polynomial is
> . P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x
u^2 + u^3f)
> . and your attempts at confusing the issue by trying to
push up x,
> . donÕt help you here, as the gÕs as factors 
of m,
necessarily have
> . a constant term with respect to m, and your claim that it
can vary
> . with m, is nonsensical on its face.
> It makes sense and I can understand it but I have to work
at it.
> I suggest that shorter sentences would make it easier for
the
> reader.
> Penny Hassett
I like it Penny Hassett, and IÕm willing to do some work, 
but
not much
as IÕm the engine that drive everything anyway.
You prepare to do some work--and make no mistake you WILL
work very
hard--and IÕll try to give you success.
After all, IÕm already one of the most powerful men on the
planet.
With your help, I can get organized and maybe do some good in
this
world.
Email me if youÕre interested, all offers are rescinded if
you do not.
James Harris
===
Subject: Re: Quick Math Guide to core error issues
>>James,
>>You post so much material that I donÕt have the time or
desire to
>>read everything and I lose track of your position and
arguments.
>>Some time ago you had an organised website with definitions
and cross-
>>reference and you have posted several summaries and
histories on this
>>newsgroup which would be easier to refer to if they were
permanently
>>available. I, for one, would be pleased if you would
resurrect your
>>website and put the information back.
> I like you Penny Hasset and appreciate your commentary
which is why
> IÕm posting in a thread where I donÕt need 
to post as itÕs
a quide.
> My problem is that I donÕt want to use MSN Groups, and I
donÕt feel
> like going to another website provider.
> I *am* willing to allow someone else to host my work as
long as they
> give me complete editorial control.
> Otherwise, itÕs easier for me to just post rather than try
to maintain
> a website.
http:www.sphosting.com is easy to use, and allows file
uploading. Of
course, if you arenÕt willing to do any work, by all means
keep everyone
somewhat confused and off-balance.
Perhaps you could post a weekly update?
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Quick Math Guide to core error issues
> <snip I like it Penny Hassett, and IÕm willing to do some
work, but not much
> as IÕm the engine that drive everything anyway.
> You prepare to do some work--and make no mistake you WILL
work very
> hard--and IÕll try to give you success.
> After all, IÕm already one of the most powerful men on the
planet.
> With your help, I can get organized and maybe do some good
in this
> world.
> Email me if youÕre interested, all offers are rescinded if
you do not.
> James Harris
of you material up on a web-site then IÕm willing to advise
on the
format and style by way of the sci.math newsgroup but thatÕs
all.
PS. Lest you feel I am being dishonest when you find out
later, let
me say that it will be obvious to many people in Britain that
I am using a nom-de-keyboard.
===
Subject: Re: Quick Math Guide to core error issues
> <snip
> I like it Penny Hassett, and IÕm willing to do some work,
but not much
> as IÕm the engine that drive everything anyway.
>
> You prepare to do some work--and make no mistake you WILL
work very
> hard--and IÕll try to give you success.
>
> After all, IÕm already one of the most powerful men on the
planet.
>
> With your help, I can get organized and maybe do some good
in this
> world.
>
> Email me if youÕre interested, all offers are rescinded if
you do not.
>
> James Harris
> of you material up on a web-site then IÕm willing to 
advise
on the
> format and style by way of the sci.math newsgroup but
thatÕs all.
last couple of days, partly out of EXTREME FRUSTRATION at my
situation. IÕve found that I can have fun with postings,
which makes
me feel better.
Still I was sincere about the $250k but am now relieved that
you
declined.
Oh yeah, IÕve taken your advice though as IÕm 
using only m as
a
variable in my recent postings as IÕm *really* ready to 
finish
things
up.
> PS. Lest you feel I am being dishonest when you find out
later, let
> me say that it will be obvious to many people in Britain
that
> I am using a nom-de-keyboard.
Oh hey, IÕd started calling you my money penny too. Kind of
like a
James thing, you know, Bond, James Bond.
Oh well, maybe someday youÕll have reason to give me your
name, but
itÕs not a big deal. In any event, unlike with Nora Baron, I
wonÕt
put quotes around your name.
James Harris
===
Subject: Re: Quick Math Guide to core error issues
> IÕm willing to do some work, but not much
I knew it! YouÕre just too lazy!
===
Subject: Re: Quick Math Guide to core error issues
putting aside the question of wether there is *any* real JSH,
I still have to question this oneÕs entitlement to the name,
when he goes over the top and says that a correspondent can
be proved to be in his newfound ring of what ever.
back to teh question of the real one:
if heÕs not being paid to do this, or has a pension
that is allowing him to make fun of Whitey (?),
then it really is pretty strange.
note on Anglo-american history:
slavery was a British institution;
thatÕs why they supported the Confederacy (along
with the New York Times etc.) with ships & materiel,
and actually organized the Civil War. (this makes
for a good, revisionist question:
What was our 3rd war with Great Britain?)
[see http://tarpley.net]
> IÕm posting in a thread where I donÕt need 
to post as itÕs
a quide.
> An explanation of what an algebraic integer is , some
> examples and their general properties would be useful.
> I vaguely recall that you had some software that would
> take a monic quadratic equation with integer coefficients
> and factorise it into linear terms with algebraic integer
> coefficients. This could usefully go here as a Java applet.
> You see, IÕve found and dropped more mathematics than most
people
> Actually *you* are in it Penny Hasset, as the object ring
is rather
> large.
> You see, you are a mathematical object which I can prove
using some
> rather basic logic and GoedelÕs proof.
> I like it that youÕre in Britain.
> I lose track of the arguments and how they are settled or
even
> if they are. Your executive summary below is a good start
but that
> is all it is. Hotlinks to expansions, with proofs, would be
better.
> If youÕre already powerful, hell, why not just do it?
> It would take great courage but if you could also document
points
> where you are no longer holding a former position then this
also
> would interesting. IÕm thinking in particular of whether a
ring
> must be closed under an infinite summation and whether Z[pi]
is
> isomorphic to the reals.
> . I already have as the polynomial is
> .
> . P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x
u^2 +
u^3f)
> .
> . and your attempts at confusing the issue by trying to
push up x,
> . donÕt help you here, as the gÕs as factors 
of m,
necessarily have
> . a constant term with respect to m, and your claim that it
can vary
> . with m, is nonsensical on its face.
--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...
La Troi Phases dÕExploitation de la Protocols des Grises de
Kyoto:
(FOSSILISATION [McCainanites?] (TM/sic))/
BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm.
Http://www.tarpley.net/bushb.htm (content partiale, below):
17 -- LÕATTEMPTER de COUP DÕETAT, 3/30/81
===
Subject: Re: Quick Math Guide to core error issues
> [...]
> note on Anglo-american history:
> slavery was a British institution;
The British slave traders bought their slaves from Arab and
African
trade slavers, so it wasnÕt a British institution it was an
international one. Slavery persists to this day of course.
> [...]
--
G.C.
===
Subject: Re: Quick Math Guide to core error issues
> For those of you trying to keep up with the mathematical
facts in the
> discussions about the error in core mathematics from a
problem with a
> definition, this post will outline the important ones
quickly and
> succinctly.
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
This is sheer idiocy.
A definition cannot lead to a contradiction.
A definition, that is not correctly understood by a wannabe
maths genius
like yourself, followed by some ridiculously confused
attempts at
proving things, can lead a sufficiently stupid person to
thinking there
are contradictions. But that is _your_ problem, and not a
problem of the
definition.
===
Subject: Re: Quick Math Guide to core error issues
In sci.physics, James Harris
<jstevh@msn.com For those of you trying to keep up with the
mathematical facts in the
> discussions about the error in core mathematics from a
problem with a
> definition, this post will outline the important ones
quickly and
> succinctly.
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
And these numbers are ... what? Presumably, you can produce
a counterexample.
> 2. The important tool I use is a polynomial:
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3 f)
> The form of the polynomial allows me to factor P(m) into
> non-polynomial factors, and the factorization with those
factors is
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> where the aÕs are roots of the following cubic:
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
Pedant point: ITYM a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2
f^2 + 3m) = 0.
> 3. Dispute centers around what happens when I divide P(m)
by f^2,
> which youÕll note is a factor of the polynomial in the 
ring
of
> algebraic integers.
That it is, for what itÕs worth. However, 
youÕve not gotten
around the f^(2/3) problem yet. I posit that a perfectly
valid transformation of your cubic is
Q(m) = P(m) / f^2 = (b_1 x + uf^(1/3))(b_2 x + uf^(1/3))(b_3
x + uf^(1/3))
where b_{i} = a_{i} / f^(2/3).
In fact, thatÕs probably what youÕd end up 
with anyway! :-)
I, however, make no claims regarding the b_{i} being
integers, algebraic or otherwise. IÕm not sure what
can be deduced therefrom.
ItÕs worth noting that f^(1/3) is an algebraic integer if f
is.
Another transformation is
R(m) = P(m) / f^3 = (c_1 x + u)(c_2 x + u) (c_3 x + u)
although in this case we have a problem, as not all the cÕs
are algebraic integers; thereÕs a missing factor of 1/f in
there somewhere. This factor can be added in:
Q(m) = f R(m) = (c_1 x + u) (c_2 x + u) (c_3 fx + uf)
but it could equally easily be added in:
Q(m) = (c_1 f^(1/3) x + u f^(1/3)) (c_2 f^(1/3) x + u f^(1/3))
(c_3 f^(1/3)x + uf^(1/3))
or
Q(m) = (c_1 x + u) ( c_2 f^(1/2) x + u f^(1/2) ) (c_3 f^(1/2)
x + u
f^(1/2))
I just donÕt know at this point.
> 4. Mathematicians have argued that f^2 divides off as a
function of m
> because if they concede that it divides off independent of
m, then I
> can show that only two of the roots of
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)
> have f as a factor.
If one sets f = 2, m = 1 we get
a^3 + 9a^2 - 28
which has as one root a_1 = -2. Factoring, we get
a^3 + 9a^2 - 28 = (a + 2) (a^2 + 7a - 14)
so the other two roots are a_x = (-7 ± sqrt(49 + 56)) / 2
= -7/2 ± sqrt(105) / 2.
Only one of these roots (-2) is divisible by 2. The other
two roots
-7/2 ± sqrt(105) / 2
are such that, if we set b_x = a_x/2, or a_x = 2b_x, we get
b_x = -7/4 ± sqrt(105) / 4.
What equation of integer coefficients does the 
bÕs satisfy?
ThatÕs simple enough; substituting 2b for a, we get
4b^2 + 14b - 14 = 0
or
2b^2 + 7b - 7 = 0.
Clearly, the bÕs are not algebraic integers, and therefore
two of the original aÕs are not divisible by 2, for this
particular setting of f and m.
This is a counterexample to your original proposition.
> 5. However, it turns out that if you go to the field of
algebraic
> numbers you can prove that for *certain* values of m and f,
the roots
> of the cubic do not have f as a factor *in the ring of
algebraic
> numbers* which is the inconsistency.
> That is, for the math to be consistent, two of the roots
*should* have
> f as a factor as long as m and f are algebraic integers,
but while I
> can show they do for a particular values like m=1,
f=sqrt(2), there
> are other values you can show they do not *in the ring of
algebraic
> integers* which results from the definition and its focus on
monic
> polynomials.
I take it you want to include
-7/4 ± sqrt(105) / 4
in the ring of algebraic integers?
Please clarify.
> Note: In the ring of algebraic integers you canÕt see the
problem but
> have to go to the field of algebraic numbers as from within
the ring
> of algebraic integers it appears that only two of the roots
have a
> factor that is f.
2/3 can be divided by 3 (the result being 2/9). Did you have a
point here?
> James Harris
--
#191, ewill3@earthlink.net
ItÕs still legal to go .sigless.
===
Subject: Re: Quick Math Guide to core error issues
> In sci.physics, James Harris
> <jstevh@msn.com>
...
> 5. However, it turns out that if you go to the field of
algebraic
> numbers you can prove that for *certain* values of m and f,
the roots
> of the cubic do not have f as a factor *in the ring of
algebraic
> numbers* which is the inconsistency.
>
> That is, for the math to be consistent, two of the roots
*should* have
> f as a factor as long as m and f are algebraic integers,
but while I
> can show they do for a particular values like m=1,
f=sqrt(2), there
> are other values you can show they do not *in the ring of
algebraic
> integers* which results from the definition and its focus on
monic
> polynomials.
> I take it you want to include
> -7/4 ± sqrt(105) / 4
> in the ring of algebraic integers?
No, he can not add both. Their sum is -7/2, and adding both
would make
2 a unit. He wants to add only one, but he will not tell us
which one.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Quick Math Guide to core error issues
In sci.physics, Dik T. Winter
<Dik.Winter@cwi.nl>
<Hn1052.19D@cwi.nl>:
> In sci.physics, James Harris
> <jstevh@msn.com ...
> 5. However, it turns out that if you go to the field of
algebraic
> numbers you can prove that for *certain* values of m and f,
the roots
> of the cubic do not have f as a factor *in the ring of
algebraic
> numbers* which is the inconsistency.
>
> That is, for the math to be consistent, two of the roots
*should*
have
> f as a factor as long as m and f are algebraic integers,
but while I
> can show they do for a particular values like m=1,
f=sqrt(2), there
> are other values you can show they do not *in the ring of
algebraic
> integers* which results from the definition and its focus on
monic
> polynomials.
>
> I take it you want to include
> -7/4 ± sqrt(105) / 4
> in the ring of algebraic integers?
> No, he can not add both. Their sum is -7/2, and adding both
would make
> 2 a unit. He wants to add only one, but he will not tell us
which one.
ItÕs a package deal. :-) And he gets 4 for the price of 2; 
the
reciprocals need to be added as well, as unit * unit = unit
and unit / unit = unit.
In fact, a lot more will be dragged in by this inclusion. But
the original definition discriminates against this number
(and for good reason).
--
#191, ewill3@earthlink.net
ItÕs still legal to go .sigless.
===
Subject: Re: Quick Math Guide to core error issues
> For those of you trying to keep up with the mathematical
facts in the
> discussions about the error in core mathematics from a
problem with a
> definition, this post will outline the important ones
quickly and
> succinctly.
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
The definition just defines a set of numbers. The 
definition
itself cannot produce a contradiction. Contradictions are
produced when one ŌtheoremÕ contradicts another 
theorem.
If you think you have a contradiction here, what is the
known theorem in algebraic number theory which is being
contradicted? (See below at {###] for my speculation on this.)
> 2. The important tool I use is a polynomial:
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3 f)
> The form of the polynomial allows me to factor P(m) into
> non-polynomial factors, and the factorization with those
factors is
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> where the aÕs are roots of the following cubic:
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
... and the aÕs are therefore algebraic integers (this cubic
is monic).
> 3. Dispute centers around what happens when I divide P(m)
by f^2,
> which youÕll note is a factor of the polynomial in the 
ring
of
> algebraic integers.
> 4. Mathematicians have argued that f^2 divides off as a
function of m
> because if they concede that it divides off independent of
m, then I
> can show that only two of the roots of
>[*] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)
> have f as a factor.
No - we donÕt argue that it divides off as a function of
m. We argue essentially that f^2 is distributed among the
factors (ai*x + u*f) in a way which depends on m. For example,
if m is such that the polynomial [*] that you give above is
reducible, then one factor is relatively prime to f. But if
m is such that [*] is irreducible, then *none* of the factors
are relatively prime to f. In fact in the irreducible case,
ALL of the factors (ai*x + u*f) are divisible by f^{2/3}.
See my post of Oct 18 in the thread Finishing argument - core
error proven for a proof of this.
> 5. However, it turns out that if you go to the field of
algebraic
> numbers you can prove that for *certain* values of m and f,
the roots
> of the cubic do not have f as a factor *in the ring of
algebraic
> numbers* which is the inconsistency.
The first part is true: it happens whenever [*] is 
irreducible,
which is true for most values of m. But it does not result
in a contradiction. The factorization is different when the
polynomial [*] is irreducible than when it is not. The twain
do not meet (i.e., [*] is either irreducible or it is not)
so there is no inconsistency.
> That is, for the math to be consistent, two of the roots
*should* have
> f as a factor as long as m and f are algebraic integers,
but while I
> can show they do for a particular values like m=1,
f=sqrt(2), there
> are other values you can show they do not *in the ring of
algebraic
> integers* which results from the definition and its focus on
monic
> polynomials.
f = sqrt(2) is of no interest here. You original concern,
relevant
not only to your claims in Advanced Polynomial Factorization
and Core
error but also to your proof of FermatÕs last theorem, dealt
with f a
a prime > 3 and m an integer relatively prime to f. Our
counterexamples
to your argument are restricted to the latter conditions. But
even if one
wanted to generalize for formal academic reasons: how f^2
distributes among
the factors (a1*x + u*f) differs as described above for
different
combinations of m and f. Proving something about the form of
the
factorization for one combination does not prove it for
others, as
you inexplicably seem to believe.
Similarly proving that for m = 0, f^2 distributes into the 3
factors as f, f, and 1, tells you nothing about cases for
which
m <> 0. I am surprised to see that you have mentioned your
erroneous
belief to the contrary, for the thousandth time.
> Note: In the ring of algebraic integers you canÕt see the
problem but
> have to go to the field of algebraic numbers as from within
the ring
> of algebraic integers it appears that only two of the roots
have a
> factor that is f.
A bizarre statement. In the field of algebraic numbers, every
number has f as a factor! This is of no interest at all. The
whole point of what you have been doing is lost if you decide
to
start talking about factorizations in a field.
Nora B.
> James Harris
[###]
Your result, if true, would contradict one of the
following theorems:
1. Roots of non-monic primitive irreducible polynomial with
integer coefficients cannot be algebraic integers.
2. The set of algebraic numbers constitutes a ring.
I think you are refusing to say that your result contradicts
1. because you have gone on record (in 2002) as accepting that
1. is a correct theorem. You have not thought much about 2.,
which is also a theorem and one which is moderately difficult
to
prove. You have chosen to say that your result follows from
an error in the definition of algebraic integers because you
know that would lead to the conclusion that mathematics is
inconsistent (which you and the rest of us abhor), OR that
your own proof is wrong. And your emotional state is such
that you cannot possibly accept the latter conclusion. But
saying that your result is a consequence of an erroneous
definition makes no sense at all on any scale.
N.B.
===
Subject: Re: Quick Math Guide to core error issues
Adjunct Assistant Professor at the University of Montana.
[.snip.]
>> 4. Mathematicians have argued that f^2 divides off as a
function of m
>> because if they concede that it divides off independent of
m, then I
>> can show that only two of the roots of
>>[*] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)
>> have f as a factor.
> No - we donÕt argue that it divides off as a function of
>m. We argue essentially that f^2 is distributed among the
>factors (ai*x + u*f) in a way which depends on m. For
example,
>if m is such that the polynomial [*] that you give above is
>reducible, then one factor is relatively prime to f.
This is not true in general either. If m=1 and f=2, one
factor is a
multiple of f, and the other two are multiples of proper
factors of f;
the polynomial [*] in that case factors as a product of a
linear and
an irreducible quadratic.
What we have said is that:
> But if
>m is such that [*] is irreducible, then *none* of the factors
>are relatively prime to f.
But there have been no general conclusions about the
reducible case in
general, other than it may indeed be the case that one of the
factors
is coprime to f.
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A manÕs capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A
great
many people are staggered to this extent, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Quick Math Guide to core error issues
...
> 1. First the problematic definition:
>
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
>
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
> The definition just defines a set of numbers. 
The definition
> itself cannot produce a contradiction. Contradictions are
> produced when one ŌtheoremÕ contradicts 
another theorem.
> If you think you have a contradiction here, what is the
> known theorem in algebraic number theory which is being
> contradicted? (See below at {###] for my speculation on
this.)
...
> Your result, if true, would contradict one of the
> following theorems:
> 1. Roots of non-monic primitive irreducible polynomial with
> integer coefficients cannot be algebraic integers.
> 2. The set of algebraic numbers constitutes a ring.
You mean algebraic integers here.
> I think you are refusing to say that your result contradicts
> 1. because you have gone on record (in 2002) as accepting
that
> 1. is a correct theorem. You have not thought much about 2.,
> which is also a theorem and one which is moderately
difficult to
> prove.
Actually he is also on record accepting that 2 is true, the
last time
was not so long ago.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Quick Math Guide to core error issues
> For those of you trying to keep up with the mathematical
facts in the
> discussions about the error in core mathematics from a
problem with a
> definition, this post will outline the important ones
quickly and
> succinctly.
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
> The definition just defines a set of numbers. 
The definition
> itself cannot produce a contradiction. Contradictions are
> produced when one ŌtheoremÕ contradicts 
another theorem.
> If you think you have a contradiction here, what is the
> known theorem in algebraic number theory which is being
> contradicted? (See below at {###] for my speculation on
this.)
> 2. The important tool I use is a polynomial:
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3 f)
> The form of the polynomial allows me to factor P(m) into
> non-polynomial factors, and the factorization with those
factors is
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> where the aÕs are roots of the following cubic:
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
> ... and the aÕs are therefore algebraic integers (this 
cubic
> is monic).
> 3. Dispute centers around what happens when I divide P(m)
by f^2,
> which youÕll note is a factor of the polynomial in the 
ring
of
> algebraic integers.
> 4. Mathematicians have argued that f^2 divides off as a
function of m
> because if they concede that it divides off independent of
m, then I
> can show that only two of the roots of
>[*] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)
> have f as a factor.
> No - we donÕt argue that it divides off as a function of
> m. We argue essentially that f^2 is distributed among the
> factors (ai*x + u*f) in a way which depends on m. For
example,
> if m is such that the polynomial [*] that you give above is
> reducible, then one factor is relatively prime to f. But if
> m is such that [*] is irreducible, then *none* of the
factors
> are relatively prime to f. In fact in the irreducible case,
> ALL of the factors (ai*x + u*f) are divisible by f^{2/3}.
> See my post of Oct 18 in the thread Finishing argument -
core
> error proven for a proof of this.
> 5. However, it turns out that if you go to the field of
algebraic
> numbers you can prove that for *certain* values of m and f,
the roots
> of the cubic do not have f as a factor *in the ring of
algebraic
> numbers* which is the inconsistency.
> The first part is true: it happens whenever [*] is
irreducible,
> which is true for most values of m. But it does not result
> in a contradiction. The factorization is different when the
> polynomial [*] is irreducible than when it is not. The twain
> do not meet (i.e., [*] is either irreducible or it is not)
> so there is no inconsistency.
> That is, for the math to be consistent, two of the roots
*should* have
> f as a factor as long as m and f are algebraic integers,
but while I
> can show they do for a particular values like m=1,
f=sqrt(2), there
> are other values you can show they do not *in the ring of
algebraic
> integers* which results from the definition and its focus on
monic
> polynomials.
> f = sqrt(2) is of no interest here. You original concern,
relevant
> not only to your claims in Advanced Polynomial
Factorization and
Core
> error but also to your proof of FermatÕs last theorem,
dealt with f
a
> a prime > 3 and m an integer relatively prime to f. Our
counterexamples
> to your argument are restricted to the latter conditions.
But even if
one
> wanted to generalize for formal academic reasons: how f^2
distributes
among
> the factors (a1*x + u*f) differs as described above for
different
> combinations of m and f. Proving something about the form
of the
> factorization for one combination does not prove it for
others, as
> you inexplicably seem to believe.
> Similarly proving that for m = 0, f^2 distributes into the 3
> factors as f, f, and 1, tells you nothing about cases for
which
> m <> 0. I am surprised to see that you have mentioned your
erroneous
> belief to the contrary, for the thousandth time.
> Note: In the ring of algebraic integers you canÕt see the
problem but
> have to go to the field of algebraic numbers as from within
the ring
> of algebraic integers it appears that only two of the roots
have a
> factor that is f.
> A bizarre statement. In the field of algebraic numbers, 
every
> number has f as a factor! This is of no interest at all. The
> whole point of what you have been doing is lost if you
decide to
> start talking about factorizations in a field.
> Nora B.
> James Harris
> [###]
> Your result, if true, would contradict one of the
> following theorems:
> 1. Roots of non-monic primitive irreducible polynomial with
> integer coefficients cannot be algebraic integers.
> 2. The set of algebraic numbers constitutes a ring.
> I think you are refusing to say that your result contradicts
> 1. because you have gone on record (in 2002) as accepting
that
> 1. is a correct theorem. You have not thought much about 2.,
> which is also a theorem and one which is moderately
difficult to
> prove. You have chosen to say that your result follows from
> an error in the definition of algebraic integers because you
> know that would lead to the conclusion that mathematics is
> inconsistent (which you and the rest of us abhor), OR that
> your own proof is wrong. And your emotional state is such
> that you cannot possibly accept the latter conclusion. But
> saying that your result is a consequence of an erroneous
> definition makes no sense at all on any scale.
> N.B.
I want to make a note about JamesÕ dividing off claim. If 
you
divide
something off, you change the equation. Let me give you an
example.
Say we have x^2-x-12=x-4.
If I solve this for x, I get x^2-2x-8=0 => (x-4)(x+2)=0 =>
x=4 or x=-2.
Well, both solutions check, but letÕs do this a little
differently, which
is
wrong.
I see that this can be written as (x-4)(x+3)=x-4.
If I divide both sides by x-4, I get x+3=1, so x=-2. Well,
this is right,
but IÕm missing a solution, the case when x-4=0.
So instead of dividing something off, you should factor it
out so you donÕt
change anything.
David Moran
===
Subject: Re: Quick Math Guide to core error issues
> 3. Dispute centers around what happens when I divide P(m)
by f^2,
> which youÕll note is a factor of the polynomial in the 
ring
of
> algebraic integers.
> 4. Mathematicians have argued that f^2 divides off as a
function of m
> because if they concede that it divides off independent of
m, then I
> can show that only two of the roots of
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)
> have f as a factor.
Yup.
> 5. However, it turns out that if you go to the field of
algebraic
> numbers you can prove that for *certain* values of m and f,
the roots
> of the cubic do not have f as a factor *in the ring of
algebraic
> numbers* which is the inconsistency.
This is bogus. As the algebraic numbers form a field, f is a
factor of
*everything*. You can not prove that something is not a
factor of
something else in a field. So if you prove that in the
algebraic numbers
something (!= 0) is not a factor of something else you only
prove that
the algebraic numbers do not form a field. In the ring of
algebraic
integers, indeed, you can prove that for certain values of m
and f,
the roots of the cubic do not have f as a factor *in that
ring*. But
that is not a inconsistency.
> Note: In the ring of algebraic integers you canÕt see the
problem but
> have to go to the field of algebraic numbers as from within
the ring
> of algebraic integers it appears that only two of the roots
have a
> factor that is f.
That is a fata morgana.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Quick Math Guide to core error issues
now, youÕve done it;
sent me on at least two, intermiable webhunts!
http://www.biblecodedigest.com/
--les ducs dÕEnron!
===
Subject: Re: Quick Math Guide to core error issues
> For those of you trying to keep up with the mathematical
facts in the
> discussions about the error in core mathematics from a
problem with a
> definition
Firstly, it is more of a diatribe by JSH, repeated in many
guises,
than a discussion, since JSH ignores most reasonable
responses and
merely repeats his refuted arguments.
Secondly, JSH is the only one claiming an error in core
mathematics. The rest of the world seems quite satisfied with
the
reliability of of mathematics, core and all.
The error JSH alleges obviously being, to anyone who has
followed
JSHÕs sequence of diatribes, that others do not see JSH as
having
the genius he sees in himself. A common problem among those of
little talent and excessive ego.
===
Subject: Re: Quick Math Guide to core error issues
> For those of you trying to keep up with the mathematical
facts in the
> discussions about the error in core mathematics from a
problem with a
> definition, this post will outline the important ones
quickly and
> succinctly.
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
Forget about the definition. Forget about the algebraic
integers.
LetÕs call them roots of monic polynomials with integer
coefficients
instead, ok?
Can you clearly formulate whatÕs wrong with roots of monic
polynomials
with integer coefficients?
===
Subject: Re: Quick Math Guide to core error issues
> For those of you trying to keep up with the mathematical
facts in the
> discussions about the error in core mathematics from a
problem with a
> definition, this post will outline the important ones
quickly and
> succinctly.
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
> 2. The important tool I use is a polynomial:
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3 f)
Yes, and then you claim you can find the terms which are
independent of
m by setting m=0. This is a blunder. Setting m=0 merely
*evaluates* P(m)
for the case m=0. Any multi-variable expression can be
simplified by
setting one or more of the variables to a constant value, of
course, but
whether you set m=0, m=1, m=I, m=Pi, or any other constant
value, you have
only succeeded in performing an evaluation of P(m) for that
value.
Naturally ŌmÕ will not appear in the resulting 
expression
because it has
been replaced by a constant. This says nothing whatsoever
about the
properties of the original expression.
For example, suppose we have:
Q(m) = 2*(m-1) + 1
Does setting m=0 remove the terms dependent on 
ŌmÕ? Is Q(m)
dependent on
ŌmÕ? (Answer: yes, it is a function of 
ŌmÕ. It has a distinct
value for
every value of ŌmÕ.)
What if:
Q(m) = 2^m + 1
does setting m=0 remove the terms dependent on 
ŌmÕ?
Suppose:
Q(m) = a^m + b*m + c/m + d*cos(m) + e*exp(m)
What about this case? Can we set m=0 to remove terms
dependent on ŌmÕ? If
not, why donÕt you explain what you mean when you state that
you can find
terms independent of ŌmÕ by placing m=0?
Apparently you either misunderstand the process of evaluating
a
multivariable expression by substitution, or you are simply
misrepresenting your argument.
--
A fool and his proof are soon refuted.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: Quick Math Guide to core error issues
> For those of you trying to keep up with the mathematical
facts in the
> discussions about the error in core mathematics from a
problem with a
> definition, this post will outline the important ones
quickly and
> succinctly.
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
> 2. The important tool I use is a polynomial:
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3 f)
Lost me again. You were doing fine in #1. Now please explain
in simple
English where youÕre going with this, before piling on the
equations.
What kind of number is excluded? If you start with the
algebraic
integers and apply standard ring operations of multiplication
and
addition, do you end up with something thatÕs not an
algebraic integer?
Or is the extra number the result of some operation other than
multiplication and addition?
Before diving into the equations, just say where youÕre 
going.
===
Subject: a statistic question
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9JEVMi16484;
I have a statistic question which I cannot solve....Hope
sombody can
help me:)
John scored 60 marks in Mathematics and 75 marks in
Programming.
The overall performance of his class in this two subjects are
: for
Mathematics, the mean score is 45 marks with a standard
deviation
of 15 marks and for Programming, the class mean is 70 with a
standard
deviation of 10 marks. In which subject did John perform
better,
Mathematics or Programming? Provide an explantion for your
answer.
===
Subject: Re: a statistic question
>John scored 60 marks in Mathematics and 75 marks in
Programming.
>The overall performance of his class in this two subjects
are : for
>Mathematics, the mean score is 45 marks with a standard
deviation
>of 15 marks and for Programming, the class mean is 70 with a
standard
>deviation of 10 marks. In which subject did John perform
better,
>Mathematics or Programming? Provide an explantion for your
answer.
Compute the z scores (might be called standard scores in your
textbook).
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Foundations broken, core error affects you all
Whether anyone wants to admit it or not, the fact that the
definition
of algebraic integers is ßawed affects everyone in the
mathematics
field.
The problem is that mathematics necessarily depends on a
correct
foundation.
My finding not only reveals a current ßaw but it also 
brings
into
question whether or not mathematicians can ever be certain
that
theyÕve covered all the bases.
ItÕd be less of a problem if mathematicians 
hadnÕt strangely
decided
to deny a result so easily provable using rather basic
mathematics,
which brings up a cultural question: Is math society
corrupted in some
strange way?
My answer is, yes, math society has clearly become corrupted
or itÕd
never sit still and leave an error at the heart of the
discipline.
Some of you may doubt that there is an error but there is and
in fact
IÕve gone so far as to have communicated in some way about 
it
with
Barry Mazur, Granville, and Ralph McKenzie--all leading
mathematicians.
They did not behave appropriately.
Now IÕm challenged with trying to figure out a 
way to get the
truth
known as it would definitely not suck to be known for the
discoveries
IÕve made.
That will involve some rather careful postings designed to
elicit a
particular response which may make some of you uncomfortable.
But remember, the people IÕm focusing on have placed
themselves
outside of the rules of society, and are acting against its
interests.
And it looks like IÕll have to go somewhat outside of where
IÕd like
to go, in order to chase them down.
James Harris
59signing consent. He is banned from virtually all the
shopping malls
in his community because he stalks young people and sexually
harasses
them. He has an extensive arrest record which includes sexual
molestation charges. HeÕs been hospitalized in mental
institutions
about his contact with young girls in many posts. Search
newsgroup
archives for posts by him containing the word nubile. As part
of his
harrassment, he provides personal details in a public forum,
such as
the real names of real children, in these and other posts.
About one
wanted her and her sister dead.
He not only curses children and prays for their death in his
posts, he
also enjoys attending the funerals of young people: And so,
since
nubile sweeties are found in greatest abundance at the
funerals of
high school students, then it is the funerals of high school
students
that make the very very best funerals, especially if there is
food...
I stuff my face (and my pockets) with all the good food and
look at
all the pretty nubile sweeties and have the time of my life..
r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%
40twister.socal.rr.
com&rnum=1
Many of his posts are sent to alt.teens.advice. However, he
liberally
offensive
missives to countless newsgroups. Some people HAVE problems
and some
folks ARE problems. DonÕt dismiss Mr. Kabatoff as a harmless
nut. When
he sends these posts to any newgroup, please help by
reporting him to
I knew of him when I was attending the University of
Saskatchewan.
HeÕd hang out in the Arts computer lab and all 
youÕd see is
screens of
numbers racing by on his laptop. I have an original copy of
his
Collecting Mail for the Coming Anti-Christ pamphlet, and have
seen
him be hauled away by campus security on more than one
occasion. My
friends and I refer to him as Crazy Number Man.
IÕve been posting to (and about) Shawn for over two years
with big
gaps in between. He has seen Pi and didnÕt like it and 
didnÕt
think it
resembled him at all. (Wrong, it fits him to a tee) He 
doesnÕt
have
total recall and has stated that he travels with a lap top to
notate
items. Also, he uses cut nÕ paste a lot if you read all the
way
through his ramblings. He is anti-social as shown by his angry
statements towards those who, by his own admission, have been
kind
(but not kind enough) to him. Still, heÕs intelligent and
seems to be
able to take a joke on occassion. ThatÕs where I came in.
ALOHA
Reply to group
(Unsolicited e-mail is deleted from the server unread
if it comes from anyone not already in my addressbook.
IÕll never even see it)
===
Subject: Core error, mathematicians, complex numbers
It turns out that focusing on one area might help break this
logjam
where apparently many of you seem to be holding out for some
way to
believe that mathematicians arenÕt trying to hide a 100+ 
year
old
error caused by the definition of algebraic integers.
Since I know physics people tend to be always use the field of
complex
numbers anyway, the following should make you feel more
comfortable.
Remember objectors have a position that requires
w_1(m) w_2(m) w_3(m) = f^2
where all the wÕs give values in the ring of algebraic
integers for
algebraic integers m and f, while varying with m.
And yes, if you did not realize thatÕs what these people 
have
been
arguing with me about for MONTHS, then you werenÕt paying
attention.
Now itÕs easy to show what they want is impossible, and 
IÕll
switch to
the field of complex numbers to use
e^{x(m)} = w_1(m), e^{y(m)} = w_2(m), e^{z(m)} = w_3(m)
so
e^{x(m) + y(m) + z(m)} = f^2
so
x(m) + y(m) + z(m) = ln f^2
and if either x(m), y(m), or z(m) has terms with m then those
terms
functions, but
an additive inverse in exponents is just a multiplicative
inverse.
ThatÕs no problem for the field of complex 
numbers, but
remember,
algebraic integers are defined to be the roots of monic
polynomials
with integer coefficients. And lots of numbers are NOT in that
ring,
like 1/2 is not.
Algebraic integers are distinctive in that, unlike complex
numbers
where every member but 0 has a multiplicative inverse,
memberd in the
ring of algebraic integer besides 0 do NOT necessarily have a
multiplicative inverse within the ring. And in fact, only
factors of
1 in the ring, i.e. units, do. And remember, to be algebraic
integers
they are roots of monic polynomials with integer coefficients,
like
x^3 + 3x + 1, for instance.
Posters have been apparently relying on the *intuitive*
notion that
still be in
the ring of algebraic integers, but that defies the results of
basic
analysis.
So to be in the ring of algebraic integers, w_1(m), w_2(m),
and w_3(m)
must be constant with respect to m, or youÕre pushed into a
field.
So *in the ring of algebraic integers* itÕs not possible to
have
w_1(m) w_2(m) w_3(m) = f^2
where all the wÕs give values in the ring of algebraic
integers for
algebraic integers m and f, while varying with m.
Now then, if youÕre naive and trusting then you may believe
that
mathematicians all over the world could make that error for
MONTHS.
James Harris
===
Subject: Re: Core error, mathematicians, complex numbers
Adjunct Assistant Professor at the University of Montana.
[.snip.]
I know youÕve retracted these comments elsewhere. I just 
want
to
comment on a little detail:
>Remember objectors have a position that requires
>w_1(m) w_2(m) w_3(m) = f^2
>where all the wÕs give values in the ring of algebraic
integers for
>algebraic integers m and f, while varying with m.
[.snip.]
>Algebraic integers are distinctive in that, unlike complex
numbers
>where every member but 0 has a multiplicative inverse,
memberd in the
>ring of algebraic integer besides 0 do NOT necessarily have a
>multiplicative inverse within the ring. And in fact, only
factors of
>1 in the ring, i.e. units, do.
ThatÕs the definition of unit: having an 
algebraic integer.
> And remember, to be algebraic integers
>they are roots of monic polynomials with integer
coefficients, like
>x^3 + 3x + 1, for instance.
>Posters have been apparently relying on the *intuitive*
notion that
still be in
>the ring of algebraic integers, but that defies the results
of basic
>analysis.
Actually, it seems to me that ->you<- have been relyin on
*intuitive*
notions. In particular, you seem to think that the
expressions for
w_1(m), etc., must in some way be given by a specific formula
into
is that you derive the conclusion that there cannot be a
function that
is not constant and has the desired properties. Your
insistence that
there could be no function that varies and has the right
properties
seems to be based on being unable to come up with some
formula that
will give it to you explicitly. Your arguments about
divisibility also
seem to betray a rather naive approach, based on intuition.
The very
use of should be a multiple seems based on hunches and
intuition.
And, as usual, you tend to project onto others what you
yourself have
done or had trouble with.
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A manÕs capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A
great
many people are staggered to this extent, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Core error, mathematicians, complex numbers
> It turns out that focusing on one area might help break
this logjam
> where apparently many of you seem to be holding out for
some way to
> believe that mathematicians arenÕt trying to hide a 100+
year old
> error caused by the definition of algebraic integers.
> Since I know physics people tend to be always use the field
of complex
> numbers anyway, the following should make you feel more
comfortable.
> Remember objectors have a position that requires
> w_1(m) w_2(m) w_3(m) = f^2
> where all the wÕs give values in the ring of algebraic
integers for
> algebraic integers m and f, while varying with m.
> And yes, if you did not realize thatÕs what these people
have been
> arguing with me about for MONTHS, then you werenÕt paying
attention.
> Now itÕs easy to show what they want is impossible,
And just as obviously, itÕs impossible to have
w_1(m) * w_2(m) = f^2
where the wÕs vary with m and are algebraic integers for
every integer m, right?
So if I claim that
w_1(m) = (m + sqrt(m^2 - 36))/2
w_2(m) = (m - sqrt(m^2 - 36))/2
vary with m, have product 9 = 3^2 for every m, and are
algebraic
integers for every integer m, then I must be lying, right?
These wÕs are the roots of x^2 - mx + 9 = 0.
By the way, w_1(0) = 3i, w_2(0) = -3i. Can you write w_1
as 3i + [part varying with m] in a way that you think
validates your thinking?
> and IÕll switch to
> the field of complex numbers to use
> e^{x(m)} = w_1(m), e^{y(m)} = w_2(m), e^{z(m)} = w_3(m)
> so
> e^{x(m) + y(m) + z(m)} = f^2
> so
> x(m) + y(m) + z(m) = ln f^2
> and if either x(m), y(m), or z(m) has terms with m then
those terms
functions,
Consider
x(m) = log[(m + sqrt(m^2 - 36))/2]
y(m) = log[(m - sqrt(m^2 - 36))/2]
with
x(m) + y(m) = log(9)
Can you identify for me the terms in m in x(m) and
in y(m)?
Now note that at m=0 we have x(0) = log[3i] = log(3)
+ i*log(pi/2) + i*2*pi*k where k=any integer.
Similarly, y(0) = log[-3i] = log(3) - i*log(pi/2) + i*2*pi*r
where r is any integer.
These values at 0 presumably have some significance to
you. You can apparently learn something from writing
x(m) as [x(m)-x(0)] + x(0), i.e. as
{log[(m + sqrt(m^2 - 36))/2]
- log(3) - i*log(pi/2) + i*2*pi*k}
+ log(3) + i*log(pi/2) + i*2*pi*k
Can you walk through this example and tell me where the
impossible part happens? Because to me it sure looks like
a simple example of what you just easily showed is
impossible.
- Randy
===
Subject: Re: Core error, mathematicians, complex numbers
> It turns out that focusing on one area might help break
this logjam
> where apparently many of you seem to be holding out for
some way to
> believe that mathematicians arenÕt trying to hide a 100+
year old
> error caused by the definition of algebraic integers.
> Since I know physics people tend to be always use the field
of complex
> numbers anyway, the following should make you feel more
comfortable.
> Remember objectors have a position that requires
> w_1(m) w_2(m) w_3(m) = f^2
> where all the wÕs give values in the ring of algebraic
integers for
> algebraic integers m and f, while varying with m.
> And yes, if you did not realize thatÕs what these people
have been
> arguing with me about for MONTHS, then you werenÕt paying
attention.
> Now itÕs easy to show what they want is impossible, and
IÕll switch to
> the field of complex numbers to use
> e^{x(m)} = w_1(m), e^{y(m)} = w_2(m), e^{z(m)} = w_3(m)
> so
> e^{x(m) + y(m) + z(m)} = f^2
> so
> x(m) + y(m) + z(m) = ln f^2
> and if either x(m), y(m), or z(m) has terms with m then
those terms
functions, but
> an additive inverse in exponents is just a multiplicative
inverse.
> ThatÕs no problem for the field of complex 
numbers, but
remember,
> algebraic integers are defined to be the roots of monic
polynomials
> with integer coefficients. And lots of numbers are NOT in
that ring,
> like 1/2 is not.
> Algebraic integers are distinctive in that, unlike complex
numbers
> where every member but 0 has a multiplicative inverse,
memberd in the
> ring of algebraic integer besides 0 do NOT necessarily have
a
> multiplicative inverse within the ring. And in fact, only
factors of
> 1 in the ring, i.e. units, do. And remember, to be
algebraic integers
> they are roots of monic polynomials with integer
coefficients, like
> x^3 + 3x + 1, for instance.
> Posters have been apparently relying on the *intuitive*
notion that
still be in
> the ring of algebraic integers, but that defies the results
of basic
> analysis.
> So to be in the ring of algebraic integers, w_1(m), w_2(m),
and w_3(m)
> must be constant with respect to m, or youÕre pushed into 
a
field.
> So *in the ring of algebraic integers* itÕs not possible 
to
have
> w_1(m) w_2(m) w_3(m) = f^2
Dik and another poster (Michael Ulm?) have given examples
like
g(x) = x^3 + m*x + f^2
That is, let w1(m), w2(m), and -w3(m) be the roots of
this monic polynomial. Then w1(m), w2(m), and
w3(m) are algebraic integers. The product of the roots is
-f^2,
so one has
w1(m)*w2(m)*w3(m) = f^2
for all algebraic integers m. An elegant, complete
counterexample. You have not yet responded to this
one.
You have been through so many arguments in the past
several days:
1. The idea that it is sufficient to consider the
terms in P(m) which are ŌindependentÕ of m,
i.e., P(0)
2. The idea that the w functions have unbounded
range, x*y = 26, etc.
3. The fill-in-the-dots idea, i.e.,
b^3 + ...+ 3(-1+m*f2)b^2 +...-(m^3 f^4 -3m^2 f^2 + 3m)
[that one was mercifully short-lived]
4. An incorrect expression for the polynomial which
the aÕs satisfy
5. A claim that b1, b2, b3 must satisfy one monic
cubic rather than two, etc.
And of course the background claim that because two
of the aÕs are divisible by f when m = 0, the same
must be true for all m.
Over and over again you are proved wrong. ItÕs
understandable that you are going to exhaust every
possible avenue before giving up. Your entire
ego is bound up in this.
The part that doesnÕt make sense is that every time
you are proved wrong, you immediately start calling
those who provided the proofs liars. They are all
liars. They said things with which you disagreed.
They were proven to be right. Therefore *they are
liars*!
How do you explain this?
Nora B.
> where all the wÕs give values in the ring of algebraic
integers for
> algebraic integers m and f, while varying with m.
> Now then, if youÕre naive and trusting then you may 
believe
that
> mathematicians all over the world could make that error for
MONTHS.
> James Harris
===
Subject: Re: Core error, mathematicians, complex numbers
> It turns out that focusing on one area might help break
this logjam
> where apparently many of you seem to be holding out for
some way to
> believe that mathematicians arenÕt trying to hide a 100+
year old
> error caused by the definition of algebraic integers.
> Since I know physics people tend to be always use the field
of complex
> numbers anyway, the following should make you feel more
comfortable.
> Remember objectors have a position that requires
> w_1(m) w_2(m) w_3(m) = f^2
> where all the wÕs give values in the ring of algebraic
integers for
> algebraic integers m and f, while varying with m.
> And yes, if you did not realize thatÕs what these people
have been
> arguing with me about for MONTHS, then you werenÕt paying
attention.
> Now itÕs easy to show what they want is impossible, and
IÕll switch to
> the field of complex numbers to use
> e^{x(m)} = w_1(m), e^{y(m)} = w_2(m), e^{z(m)} = w_3(m)
> so
> e^{x(m) + y(m) + z(m)} = f^2
And the rest doesnÕt work, as someone on sci.math pointed 
out
a good
example.
ItÕs like if you have x^2 + mx - f^2, the roots 
define
functions that
multiply to give f^2 over all of algebraic integer m.
James Harris
===
Subject: Re: Core error, mathematicians, complex numbers
> It turns out that focusing on one area might help break
this logjam
> where apparently many of you seem to be holding out for
some way to
> believe that mathematicians arenÕt trying to hide a 100+
year old
> error caused by the definition of algebraic integers.
> Since I know physics people tend to be always use the field
of complex
> numbers anyway, the following should make you feel more
comfortable.
> Remember objectors have a position that requires
> w_1(m) w_2(m) w_3(m) = f^2
> where all the wÕs give values in the ring of algebraic
integers for
> algebraic integers m and f, while varying with m.
And what the heck is wrong with that?
Consider the cubic:
x^3 + mx - f^2 = 0
The roots of this are obviously algebraic integers for all
integers m, their product is f^2, and they depend on m.
Explicitly by the Cardano formula
(http://mathworld.wolfram.com/CubicEquation.html)
define
Q = m/3
R = -f^2/2
D = Q^3 + R^2 = m^3/27 + f^4/4
S = (R + sqrt(D))^(1/3)
T = (R - sqrt(D))^(1/3)
then the three roots of this, call them w1, w2 and w3 are
w1 = S + T
w2 = [-(S+T) + i*sqrt(3)*(S-T)]/2
w3 = [-(S+T) - i*sqrt(3)*(S-T)]/2
The dependence on m is contained in
S(m) = (-f^2/2 + sqrt(m^3/27 + f^4/4))^(1/3)
and
T(m) = (-f^2/2 - sqrt(m^3/27 + f^4/4))^(1/3)
Clearly w1, w2 and w3 depend on m, they are algebraic integers
for all integers m, and w1*w2*w3 = f^2.
What does your business of setting m=0 prove? You get
S(0) = (-f^2/2 + sqrt(f^4/4))^(1/3) = 0
T(0) = (-f^2/2 - sqrt(f^4/4))^(1/3) = (-f^2)^(1/3)
= (-1)^(1/3) * f^(2/3)
w1(0) = T(0)
w2(0) = -T(0)*[1 + i*sqrt(3)]/2
w3(0) = -T(0)*[1 - i*sqrt(3)]/2
And what does that show exactly?
> And yes, if you did not realize thatÕs what these people
have been
> arguing with me about for MONTHS, then you werenÕt paying
attention.
> Now itÕs easy to show what they want is impossible, and
IÕll switch to
> the field of complex numbers to use
> e^{x(m)} = w_1(m), e^{y(m)} = w_2(m), e^{z(m)} = w_3(m)
> so
> e^{x(m) + y(m) + z(m)} = f^2
> so
> x(m) + y(m) + z(m) = ln f^2
> and if either x(m), y(m), or z(m) has terms with m then
those terms
functions,
Nope. You posted this same statement at least 3 times already,
and each time Dik Winter pointed out that the logarithm is
multi-valued in the complex numbers. There are infinitely
many x(m) such that e^(x(m)) = w_1(m), and there are
infinitely many solutions to
e^(x(m) + y(m) + z(m)) = f^2
They satisfy
x(m) + y(m) + z(m) = log(f^2) + 2*pi*k*i
where k is any integer.
- Randy
===
Subject: Re: Core error, mathematicians, complex numbers
> It turns out that focusing on one area might help break
this logjam
> where apparently many of you seem to be holding out for
some way to
> believe that mathematicians arenÕt trying to hide a 100+
year old
> error caused by the definition of algebraic integers.
> Since I know physics people tend to be always use the field
of complex
> numbers anyway, the following should make you feel more
comfortable.
> Remember objectors have a position that requires
> w_1(m) w_2(m) w_3(m) = f^2
> where all the wÕs give values in the ring of algebraic
integers for
> algebraic integers m and f, while varying with m.
> And yes, if you did not realize thatÕs what these people
have been
> arguing with me about for MONTHS, then you werenÕt paying
attention.
> Now itÕs easy to show what they want is impossible, and
IÕll switch to
> the field of complex numbers to use
> e^{x(m)} = w_1(m), e^{y(m)} = w_2(m), e^{z(m)} = w_3(m)
> so
> e^{x(m) + y(m) + z(m)} = f^2
> so
> x(m) + y(m) + z(m) = ln f^2
> and if either x(m), y(m), or z(m) has terms with m then
those terms
functions, but
> an additive inverse in exponents is just a multiplicative
inverse.
> ThatÕs no problem for the field of complex 
numbers, but
remember,
> algebraic integers are defined to be the roots of monic
polynomials
> with integer coefficients. And lots of numbers are NOT in
that ring,
> like 1/2 is not.
> Algebraic integers are distinctive in that, unlike complex
numbers
> where every member but 0 has a multiplicative inverse,
memberd in the
> ring of algebraic integer besides 0 do NOT necessarily have
a
> multiplicative inverse within the ring. And in fact, only
factors of
> 1 in the ring, i.e. units, do. And remember, to be
algebraic integers
> they are roots of monic polynomials with integer
coefficients, like
> x^3 + 3x + 1, for instance.
> Posters have been apparently relying on the *intuitive*
notion that
still be in
> the ring of algebraic integers, but that defies the results
of basic
> analysis.
> So to be in the ring of algebraic integers, w_1(m), w_2(m),
and w_3(m)
> must be constant with respect to m, or youÕre pushed into 
a
field.
> So *in the ring of algebraic integers* itÕs not possible 
to
have
> w_1(m) w_2(m) w_3(m) = f^2
> where all the wÕs give values in the ring of algebraic
integers for
> algebraic integers m and f, while varying with m.
> Now then, if youÕre naive and trusting then you may 
believe
that
> mathematicians all over the world could make that error for
MONTHS.
People have been arguing with Mr. Harris for months, to no
avail.
Clearly Mr. Harris is delusional, possibly deranged. It is not
possible to argue with such a person and that has become
evident over
the hundreds of messages posted to this newsgroup.
Mr. Ullrich used a quotation from De Morgan on the topic of
replying
to crackpots to the effect that doing so is necessary to
prevent the
gullible public at large from believing them. Unfortunately
arguing
with mentally ill people will never cease. How much valuable
mathematics might be done in the time people have taken to
respond to
Mr. Harris? How much time has been wasted on his postings?
Ignoring Mr. Harris wonÕt lead to something bad. In fact it
is the
right thing to do. It may even be the compassionate thing to
do. I
recommend the liberal use of the killfile
===
Subject: Re: Core error, mathematicians, complex numbers
> People have been arguing with Mr. Harris for months, to no
avail.
> Clearly Mr. Harris is delusional, possibly deranged. It is
not
> possible to argue with such a person and that has become
evident over
> the hundreds of messages posted to this newsgroup.
It is, however, both possible and desirable to prevent others
from
being entrapped by JSHÕs folie.
===
Subject: Re: Core error, mathematicians, complex numbers
...
> e^{x(m)} = w_1(m), e^{y(m)} = w_2(m), e^{z(m)} = w_3(m)
> so
> e^{x(m) + y(m) + z(m)} = f^2
> so
> x(m) + y(m) + z(m) = ln f^2
Does not follow.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Is zero in the domain of functions ?
Suppose that F:D_1--->R , G:D_2 --->R
F(x)=sqrt(x^2(x-1)*(x+1)) , G(x)= |x|*sqrt{(x-1)*(x+1)}
where D_1 ,D_2 are the domains of definitions , for instance
D_1 := {x in R ; F(x) is real} .
Questions: i) ItÕs true that D_1 = D_2 ?
ii) Are the above two functions equal ?
iii) The point x_0 = 0 belongs to D_j , j=1 or j=2 ?
===
Subject: Re: Is zero in the domain of functions ?
> Suppose that F:D_1--->R , G:D_2 --->R
> F(x)=sqrt(x^2(x-1)*(x+1)) , G(x)= |x|*sqrt{(x-1)*(x+1)}
> where D_1 ,D_2 are the domains of definitions , for instance
> D_1 := {x in R ; F(x) is real} .
> Questions: i) ItÕs true that D_1 = D_2 ?
Yes - see below.
> ii) Are the above two functions equal ?
Yes. But a word of caution: sqrt((-4)*(-9)) =
sqrt(-4)*sqrt(-9)?
> iii) The point x_0 = 0 belongs to D_j , j=1 or j=2 ?
Neither. D_1 = D_2 = (oo, -1] / [1. oo).
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: Is zero in the domain of functions ?
> Suppose that F:D_1--->R , G:D_2 --->R
> F(x)=sqrt(x^2(x-1)*(x+1)) , G(x)= |x|*sqrt{(x-1)*(x+1)}
> where D_1 ,D_2 are the domains of definitions , for instance
> D_1 := {x in R ; F(x) is real} .
This sort of question, about implied domains, has been asked
before in
math newsgroups.
> Questions: i) ItÕs true that D_1 = D_2 ?
> Yes - see below.
I agree. But I can imagine how some might disagree; see below.
> ii) Are the above two functions equal ?
> Yes. But a word of caution: sqrt((-4)*(-9)) =
sqrt(-4)*sqrt(-9)?
I agree. But I can imagine how some might disagree; see below.
> iii) The point x_0 = 0 belongs to D_j , j=1 or j=2 ?
> Neither. D_1 = D_2 = (oo, -1] / [1. oo).
Both, I would say. D_1 = D_2 = (-oo, -1] / {0} / [1, +oo).
As to those mentioned above who might disagree:
Some could, reasonably, argue that 0 is in the domain of F,
but not in the
domain of G.
BTW, I really canÕt imagine why Paul excludes 0 from the
domain of F.
David
===
Subject: Less symbols, core error proof
1. First the problematic definition:
Algebraic integers are defined to be roots of monic
polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17,
where
monic refers to the leading coefficient.
My assertion is that the over hundred year old definition
excludes
numbers that have to be included to keep from having
contradiction
i.e. mathematical inconsistency.
2. The important tool I use is a polynomial:
P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7)
The form of the polynomial allows me to factor P(m) into
non-polynomial factors, and the factorization with those
factors is
P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)
where the aÕs are roots of the following cubic:
a^3 + 3(-1 + 49m)a^2 - f^2(2401 m^3 - 147 m^2 + 3m).
g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7)
but setting m=0, gives me P(0) = 49(3(5) + 7), which fits with
the
cubic as at m=0 it gives
a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3,
to show that at m=0, the three factors are
g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22.
Now dividing P(m) by 49 gives
P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7
and the question is what happens to the gÕs, but look now at
P(0)/49)
as that is
P(0)/49 = 3(5) + 7
as two factors of 7, each 7, have beeen divided off, which is
easy to
see.
But 7, 7, and 49 are NOT functions of m, as they are just
numbers, so
those factors must go *independent* of the value of m, which
means
that what value of m I choose doesnÕt matter so now I can go
to the
full expression and get
P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
here it may seem that I just arbitarily divided through, but
consider
what happens if you try some other combination, like
P(m)/49 =
(5 a_1/7^{2/3} + 7^{1/3})(5 a_2/7^{2/3} + 7^{1/3})(5
a_3/7^{2/3} +
7^{1/3})
as then letting m=0 gives
P(0)/49 = 7^{1/3} 7^{1/3} (5(3)/7^{2/3} + 7^{1/3}).
While it may seem possible that the 7Õs roam around based on
the value
of m, thereÕs just no mathematical reason for them to do so
because,
well, 7 is 7, and it is NOT a function of m.
Now the problem is based on the factorization
P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
two of the aÕs *should* have 7 as a factor, and in fact they
do, in a
proper ring, but the ring of algebraic integers has problems,
so that
for certain values of m, they wonÕt, while maybe (I 
havenÕt
checked)
for some value of m, they will.
ItÕs that *inconstency* which shows you 
thereÕs a problem
because
mathematics isnÕt about being wishy-washy, where sometimes
something
works and then other times it doesnÕt.
That error has sat in mathematics, the body of discoveries
commonly
called mathematics, for over a *hundred* years.
James Harris
===
Subject: Re: Less symbols, core error proof
Let me try to follow your reasoning. I omit point 1, because
it consists
of preliminary comments only, and is just a rehash of things
that have
been written too many times already.
I try to follow this with a quadratic, using the same line of
reasoning.
> 2. The important tool I use is a polynomial:
> P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 +
7)
I start with:
Q(m) = 7((2.m^2 - 3.m).5^2 - 3.(-1 + m).5 + 7)
> The form of the polynomial allows me to factor P(m) into
> non-polynomial factors, and the factorization with those
factors is
> P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)
Q(m) = (5.a1 + 7)(5.a2 + 7)
> where the aÕs are roots of the following cubic:
> a^3 - 3(-1 + 49m)a^2 - f^2(2401 m^3 - 147 m^2 + 3m).
(ŌfÕ is a typo for 
Ō7Õ.)
a^2 + 3(-1 + m).a + 7.(2.m^2 - 3.m)
> g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7)
(the second Ōa_1Õ is a typo for 
Ōa_3Õ.)
I get two factors:
g1 = (5a1 + 7) and g2 = (5a2 + 7)
> but setting m=0, gives me P(0) = 49(3(5) + 7), which fits
with the
> cubic as at m=0 it gives
Setting m=0, gives me Q(0) = 7(3.5 + 7), which fits with the
cubic as
at m=0 it gives
> a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3,
a^2 - 3a = 0, so a1 = 0, and a2 = 3,
> to show that at m=0, the three factors are
> g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22.
g1 = 7, g2 = 3.5 + 7 = 22.
> Now dividing P(m) by 49 gives
> P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7
Dividing Q(m) by 7 gives
Q(m)/7 = (2.m^2 - 3.m).5^2 - 3.(-1 + m).5 + 7
> and the question is what happens to the gÕs, but look now
at P(0)/49)
> as that is
> P(0)/49 = 3(5) + 7
Look at Q(0)/7 as that is
Q(0)/7 = 3.5 + 7
> as two factors of 7, each 7, have beeen divided off, which
is easy to
> see.
As one factor of 7 has been divided off.
> But 7, 7, and 49 are NOT functions of m, as they are just
numbers, so
> those factors must go *independent* of the value of m,
which means
> that what value of m I choose doesnÕt matter so now I can
go to the
> full expression and get
> P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
Q(m)/7 = (5.a1/7 + 1)(5.a3 + 7)
> here it may seem that I just arbitarily divided through,
but consider
> what happens if you try some other combination, like
> P(m)/49 =
> (5 a_1/7^{2/3} + 7^{1/3})(5 a_2/7^{2/3} + 7^{1/3})(5
a_3/7^{2/3} +
> 7^{1/3})
Q(m)/7 = (5.a1/sqrt(7) + sqrt(7))(5.a2/sqrt(7) + sqrt(7))
> as then letting m=0 gives
> P(0)/49 = 7^{1/3} 7^{1/3} (5(3)/7^{2/3} + 7^{1/3}).
[ I note that this evaluates to 5.3 + 7 = 22, the correct
answer.]
Q(0)/7 = sqrt(7) (5.3/sqrt(7) + sqrt(7)) = 22.
> While it may seem possible that the 7Õs roam around based
on the value
> of m, thereÕs just no mathematical reason for them to do 
so
because,
> well, 7 is 7, and it is NOT a function of m.
> Now the problem is based on the factorization
> P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
Q(m)/7 = (5.a1/7 + 1)(5.a2 + 7)
> two of the aÕs *should* have 7 as a factor, and in fact
they do, in a
> proper ring, but the ring of algebraic integers has
problems, so that
> for certain values of m, they wonÕt, while maybe (I 
havenÕt
checked)
> for some value of m, they will.
One of the aÕs *should* have 7 as a factor, and in fact it
does, in a
proper ring, at least that would be the case according to you
reasoning.
But....
Remember the quadratic for a:
a^2 + 3(-1 + m).a + 7.(2.m^2 - 3.m)
set m = 1. We get
a^2 - 7
the roots are +- sqrt(7). Which of these two is divisible by
7? You
may note that if you have a ring where one is divisible by 7,
the other
is also divisible by 7. So in a ring where one is divisible
by 7, 7
is a unit, but in that case both are divisble by 7; in fact
all numbers
in that ring are divisible by 7.
Now please state where I am not correct.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Less symbols, core error proof
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
What part of the *definition* is inconsistent? For it to be
inconsistent would suggest that there are no algebraic
integers.
> 2. The important tool I use is a polynomial:
> P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 +
7)
> The form of the polynomial allows me to factor P(m) into
> non-polynomial factors, and the factorization with those
factors is
> P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)
> where the aÕs are roots of the following cubic:
> a^3 + 3(-1 + 49m)a^2 - f^2(2401 m^3 - 147 m^2 + 3m).
> g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7)
> but setting m=0, gives me P(0) = 49(3(5) + 7), which fits
with the
> cubic as at m=0 it gives
> a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3,
> to show that at m=0, the three factors are
> g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22.
Have you looked at P(1) = 2^2 3^2 7^2 7817?
How about P(-1) = 2 7^2 159059?
> Now dividing P(m) by 49 gives
> P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7
> and the question is what happens to the gÕs, but look now
at P(0)/49)
> as that is
> P(0)/49 = 3(5) + 7
> as two factors of 7, each 7, have beeen divided off, which
is easy to
> see.
> But 7, 7, and 49 are NOT functions of m, as they are just
numbers, so
> those factors must go *independent* of the value of m,
which means
> that what value of m I choose doesnÕt matter so now I can
go to the
> full expression and get
P(0) is *also* not a function of m. It is just a number, yet
you want
to represent itÕs factorization quite oddly as if it were a
function of m.
> P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
> here it may seem that I just arbitarily divided through,
It does look that way, yes.
> but consider
> what happens if you try some other combination, like
> P(m)/49 =
> (5 a_1/7^{2/3} + 7^{1/3})(5 a_2/7^{2/3} + 7^{1/3})(5
a_3/7^{2/3} +
> 7^{1/3})
> as then letting m=0 gives
> P(0)/49 = 7^{1/3} 7^{1/3} (5(3)/7^{2/3} + 7^{1/3}).
> While it may seem possible that the 7Õs roam around based
on the value
> of m, thereÕs just no mathematical reason for them to do 
so
because,
> well, 7 is 7, and it is NOT a function of m.
But P(m) is a function of m, while P(0) is not. Why should
they behave
the same way? Have you done any work with other values of m,
such as 1?
> Now the problem is based on the factorization
> P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
> two of the aÕs *should* have 7 as a factor, and in fact
they do, in a
> proper ring, but the ring of algebraic integers has
problems, so that
> for certain values of m, they wonÕt, while maybe (I 
havenÕt
checked)
> for some value of m, they will.
You claim they should, but that would only be the case if
P(m) and P(0)
were the same thing. They arenÕt even the same type of 
thing,
as they
exist in different rings.
Also, what does any of this have to do with the behavior of
algebraic
integers as a ring? I donÕt recall rings having any 
particular
requirements along the lines of the one you imply here.
> ItÕs that *inconstency* which shows you 
thereÕs a problem
because
> mathematics isnÕt about being wishy-washy, where sometimes
something
> works and then other times it doesnÕt.
This isnÕt an inconsistency in the mathematical sense. Your
use of
w(m)Õs removed the apparent inconsistency quite neatly,
except you donÕt
want to accept it.
> That error has sat in mathematics, the body of discoveries
commonly
> called mathematics, for over a *hundred* years.
What does any of this have to do with the definition of an
algebraic
integer? You are talking about properties that are
*consequences* of
the definition and that you donÕt seem to like 
the way they
behave.
Perhaps this says more about your expectations.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Less symbols, core error proof
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
> 2. The important tool I use is a polynomial:
> P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 +
7)
> The form of the polynomial allows me to factor P(m) into
> non-polynomial factors, and the factorization with those
factors is
> P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)
> where the aÕs are roots of the following cubic:
> a^3 + 3(-1 + 49m)a^2 - f^2(2401 m^3 - 147 m^2 + 3m).
> g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7)
Note that the gÕs and aÕs are functions of m.
> but setting m=0, gives me P(0) = 49(3(5) + 7), which fits
with the
> cubic as at m=0 it gives
> a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3,
> to show that at m=0, the three factors are
So when m=0, 5.a_1(0) + 7 is divisible by 7. Note that this
is because
a_1(0) is divisible by 7.
> g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22.
> Now dividing P(m) by 49 gives
> P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7
> and the question is what happens to the gÕs, but look now
at P(0)/49)
> as that is
> P(0)/49 = 3(5) + 7
> as two factors of 7, each 7, have beeen divided off, which
is easy to
> see.
> But 7, 7, and 49 are NOT functions of m, as they are just
numbers, so
> those factors must go *independent* of the value of m,
which means
> that what value of m I choose doesnÕt matter so now I can
go to the
> full expression and get
This is a wrong assertion. The distribution of the factors of
7 amongst
the factors of the polynomials depends on the divibility of
the aÕs.
And as the aÕs are dependent on m, so is the divisibility,
and so is
the distribution of the factors.
> P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
> here it may seem that I just arbitarily divided through,
but consider
> what happens if you try some other combination, like
> P(m)/49 =
> (5 a_1/7^{2/3} + 7^{1/3})(5 a_2/7^{2/3} + 7^{1/3})(5
a_3/7^{2/3} +
> 7^{1/3})
You remain assuming that the factors of 7 distribute
similarly for all
cases amongst the factors of the polynomial. This is false.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Less symbols, core error proof
_Fewer_ symbols, surely?
===
Subject: Simple principle in core error proof
What makes my situation especially frustrating is how simple
the
argument is that proves thereÕs this problem with algebraic
integers,
yet still I have to keep explaining.
The basic principle is like with
P(m) = 2(m^2 + 2m + 1) = (a_1 m + 2)(a_2 m + 1)
where most of you can probably guess what factor a_1 must
have!!!
Now the polynomial I use is more complicated such that I need
to set
m=0 to figure things out, but notice here what happens:
P(0) = 2, and dividing off 2 from P(m) gives
P(m)/2 = m^2 + 2m + 1
and notice that P(0)/2 = 1, which tell you that the
independent term
changed.
Given
(a_1 m + 2)(a_2 m + 1)
itÕs clear that a_1 has a factor that is 2.
ItÕs a simple idea that I use with a more complicated
polynomial, and
I think that the reason so many math people go out of their
way to
make it seem like itÕs wrong is that theyÕre 
embarrassed by
the over
hundred year old error that I found.
IÕve been surprised at how dedicated they can be at trying 
to
hide the
truth.
After all, IÕve communicated with top mathematicians like
Barry Mazur,
Granville, and Ralph McKenzie, who may be people many of you
havenÕt heard of, but in certain math circles 
theyÕre
well-known.
In McKenzieÕs case I explained it all to him ***in-person***
and he
basically blew me off.
These mathematicians are hellbent on trying to hide that their
discipline could actually have a ßaw like this for as long as
they
can get away with it.
ItÕs wrong, and it canÕt help world society.
Meanwhile itÕs not doing me a bit of good either, and I 
think
that
part of their motivation is making me miserable, knowing that
what I
found *should* get me accolades.
Yup, call me crazy, but I think the bastards are out to get
me!!!
James Harris
===
Subject: Re: Simple principle in core error proof
> What makes my situation especially frustrating is how
simple the
> argument is that proves thereÕs this problem with 
algebraic
integers,
> yet still I have to keep explaining.
Ask yourself this. How can a definition lead to a
contradiction?
Suppose I define a set B = {all objects in the universe that
are
left-handed giraffes} and then I look at every object in the
universe
and ask, is it a left-handed giraffe? If it is, itÕs in B. 
If
not,
itÕs not in B. Perhaps there are no left-handed giraffes, in
which case
B is still well-defined, it just happens to be empty.
If I define the algebraic integers as real numbers that are
the root of
a polynomial with monic leading coefficient, then I can look
at each
real number and say itÕs either in this set or 
isnÕt.
Can you explain what you mean when you say you are getting a
contradiction from the definition? For example, have you found
a real
number that both is and isnÕt an algebraic integer? What
contradiction,
exactly, have you found?
===
Subject: Re: Simple principle in core error proof
> What makes my situation especially frustrating is how
simple the
> argument is that proves thereÕs this problem with 
algebraic
integers,
> yet still I have to keep explaining.
There is a meaning of simple which applies to JSHÕs
explanation,
meaning authored by a simpleton, which JSH is mathematically
at
least.
===
Subject: Re: Simple principle in core error proof
Adjunct Assistant Professor at the University of Montana.
>What makes my situation especially frustrating is how simple
the
>argument is that proves thereÕs this problem with algebraic
integers,
>yet still I have to keep explaining.
>The basic principle is like with
>P(m) = 2(m^2 + 2m + 1) = (a_1 m + 2)(a_2 m + 1)
>where most of you can probably guess what factor a_1 must
have!!!
This is a red herring. Here, the polynomial has a coefficients
which
are constant with respect to the variable of factorization,
and what
you are varying is the polynomial variable m. In your actual
situation, you are factoring with respect to a DIFFERENT
variable, X,
and the coefficients are not constant, but are instead
functions of
m.
This example is thus worthless as a guide.
>Now the polynomial I use is more complicated such that I
need to set
>m=0 to figure things out, but notice here what happens:
What happens is you engage in a red herring. In your actual
application, your factorization is not with respect to m, but
with
respect to x. And you do not set x equal to 0, you set m
equal to
0. What you are doing here is in no way parallel to what you
do in
your actual application, and thus conclusions you draw from
this
example are not necessarily applicable to your actual
application.
[.red herring left in place so James cannot complain that I
Ōremoved the mathÕ.]
>P(0) = 2, and dividing off 2 from P(m) gives
>P(m)/2 = m^2 + 2m + 1
>and notice that P(0)/2 = 1, which tell you that the
independent term
>changed.
>Given
>(a_1 m + 2)(a_2 m + 1)
>itÕs clear that a_1 has a factor that is 2.
Because a1 and a2 are constant. They do not vary. So whatever
happens
at one point is necessarily what happens at all points. This
is not what
happens in your actual application; in your application, the
values of
the coefficients change as you change m, and it does not
follow that
what happens at one value of m is what happens at all values
of m. In
fact, it does not, has been amply demonstrated by explicit
calculation.
>ItÕs a simple idea that I use with a more complicated
polynomial,
Then you got confused. here, you have a polynomial in m being
factored
as linear terms in m, where the coefficients with respect to m
are
constant. In your application, you are factoring a polynomial
which, as
m varies, is reducible or irreducible over Q depending on the
value of
m; here, you have a polynomial which is always reducible over
Q
regardless of the value of the (non-existent) non-polynomial
variable.
In your application, you have a polynomial in m being
factored into
linear terms WITH RESPECT TO A DIFFERENT VARIABLE x, and the
coefficients with respect to m are ->functions of x<-, and the
coefficients of x are functions of m. Here you set the
polynomial
variable equal to 0, but in your application you set m, not
the
polynomial variable at issue, equal to 0. Here you KNOW the
coefficients are constant and therefore whatever happens at
one point
will happen at all points, but in your application the
coefficients
are NOT constant and therefore it does not follow that what
happens at
one point will happen at all points. The situations are NOT
similar,
and therefore the situations are not comparable.
[.snip.]
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A manÕs capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A
great
many people are staggered to this extent, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Simple principle in core error proof
>What makes my situation especially frustrating is how simple
the
>argument is that proves thereÕs this problem with algebraic
integers,
>yet still I have to keep explaining.
>The basic principle is like with
>P(m) = 2(m^2 + 2m + 1) = (a_1 m + 2)(a_2 m + 1)
>where most of you can probably guess what factor a_1 must
have!!!
> This is a red herring. Here, the polynomial has a
coefficients which
> are constant with respect to the variable of factorization,
and what
> you are varying is the polynomial variable m. In your actual
> situation, you are factoring with respect to a DIFFERENT
variable, X,
> and the coefficients are not constant, but are instead
functions of
> m.
> This example is thus worthless as a guide.
Yet readers can see the reality here:
P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7)
The form of the polynomial allows me to factor P(m) into
non-polynomial factors, and the factorization with those
factors is
P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)
where ONLY m is a variable.
And yes I used more symbols before but found that posters
like Arturo
Magidin could too easily confuse people about what was going
on, so
IÕve put in values where once there were symbols.
Unfortunately, Arturo Magidin is a rather evil person who
lies about
the math as if the core error should stay hidden.
HeÕs bad, heÕs evil, and IÕm 
sick of his crap.
James Harris
===
Subject: Re: Simple principle in core error proof
Adjunct Assistant Professor at the University of Montana.
>Unfortunately, Arturo Magidin is a rather evil person who
lies about
>the math as if the core error should stay hidden.
>HeÕs bad, heÕs evil, and IÕm 
sick of his crap.
Yet, when I offered to stop posting if you would only tell me
to do
so, you ->refused<-. Were you too scared to tell someone to
stop
posting, or are you just lying here about being sick of [my]
crap?
(No, I wonÕt count as a lie the fact that you have given no
evidence
whatsoever of anything I have posted being false, let along
ŌcrapÕ;
thatÕs just your usual lying to yourself).
Why, if as you claim I am bad, evil, rather evil, someone who
lies about the math, and if you are sick of [my] crap, did you
refuse my offer to Ōstep out of your wayÕ, as it 
were, by no
longer
posting to anything you might say?
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A manÕs capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A
great
many people are staggered to this extent, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Simple principle in core error proof
>>What makes my situation especially frustrating is how
simple the
>>argument is that proves thereÕs this problem with 
algebraic
integers,
>>yet still I have to keep explaining.
>>The basic principle is like with
>>P(m) = 2(m^2 + 2m + 1) = (a_1 m + 2)(a_2 m + 1)
>>where most of you can probably guess what factor a_1 must
have!!!
>> This is a red herring. Here, the polynomial has a
coefficients which
>> are constant with respect to the variable of
factorization, and what
>> you are varying is the polynomial variable m. In your
actual
>> situation, you are factoring with respect to a DIFFERENT
variable, X,
>> and the coefficients are not constant, but are instead
functions of
>> m.
>> This example is thus worthless as a guide.
>Yet readers can see the reality here:
>P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7)
>The form of the polynomial allows me to factor P(m) into
>non-polynomial factors, and the factorization with those
factors is
>P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)
>where ONLY m is a variable.
And can you PROVE that a_1, a_2, and a_3 will satisfy the
properties
you want under these conditions? What are those properties?
You can ALWAYS get such a factorization, if you allow a_1,
a_2, and
a_3 to be algebraic numbers; but then your congruence
arguments bite
the dust.
>Unfortunately, Arturo Magidin is a rather evil person who
lies about
>the math as if the core error should stay hidden.
Unfortunately, James Harris is a rather dishonest person who
lies about me,
having been unable to find a single instance of me lying about
the
math, and who, each time he accuses me of such behavior, ends
up
retracting his claim but not his accusation.
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A manÕs capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A
great
many people are staggered to this extent, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Problem with definition of algebraic integers
1. First the problematic definition:
Algebraic integers are defined to be roots of monic
polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17,
where
monic refers to the leading coefficient.
My assertion is that the over hundred year old definition
excludes
numbers that have to be included to keep from having
contradiction
i.e. mathematical inconsistency.
2. The important tool I use is a polynomial:
P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7)
where m varies in the ring of algebraic integers.
Some may find it looks odd. However, the form of the 
polynomial
allows me to factor P(m) into non-polynomial factors, and the
factorization with those factors is
P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)
where the aÕs are roots of the following cubic:
a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m).
g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7)
but setting m=0, gives me P(0) = 49(3(5) + 7), which fits with
the
cubic as at m=0 it gives
a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3,
to show that at m=0, the three factors are
g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22.
Now dividing P(m) by 49 gives
P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7
and the question is what happens to the gÕs, but look now at
P(0)/49)
as that is
P(0)/49 = 3(5) + 7
as two factors of 7, each 7, have beeen divided off, which is
easy to
see.
But 7, 7, and 49 are NOT functions of m, as they are just
numbers, so
those factors must go *independent* of the value of m, which
means
that what value of m I choose doesnÕt matter so now I can go
to the
full expression and get
P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
where the idea is almost trivially simple as consider a
polynomial
like
S(m) = 7(m^2 + 2m + 1) = (b_1 m + 7)(b_2 m + 1)
and notice that S(0) = 7, while S(m)/7, gives you
S(m)/7 = m^2 + 2m + 1,
which means that
S(m)/7 = (b_1 m/7 + 1)(b_2 m + 1) = m^2 + 2m + 1
while I had the more complicated
P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7
but the same principle works as just like with S(m), with P(m)
dividing out factors that are 7 affects the independent or
constant
terms, revealing factors of the aÕs and bÕs.
That is, from the distributive property, factors that are 7
must
divide through.
So now I know that the correct factorization is
P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
which is like
S(m)/7 = (b_1 m/7 + 1)(b_2 m + 1).
3. So *two* of the aÕs *should* have 7 as a factor, and in
fact they
do, in a proper ring, but the ring of algebraic integers has
problems,
so that for certain values of m, they wonÕt.
ItÕs that *inconsistency* which shows you 
thereÕs a problem
because
mathematics isnÕt about being wishy-washy, where sometimes
something
works and then other times it doesnÕt.
That error has sat in mathematics--the body of discoveries
commonly
called mathematics--for over a *hundred* years.
Since I found the error I should probably get rich and famous
from it,
but so far mathematicians IÕve contacted seem more 
interested
in
denying or hiding the error than in telling the truth.
However, that
means there is this error, which may sink lots of proofs over
the
past hundred years in an area of mathematics called algebraic
number
theory.
Some of them may be trying to hide it partly out of envy or
jealousy
of my discovery as well. Mathematicians can be VERY vicious
for petty
and childish reasons IÕve found.
If you are a math student, you probably will want to stay out
of the
area of algebraic number theory, or consider carefully things
your
professors supposedly prove in the area.
While mathematicians behave this way, you have to wonder now
about
what they teach you.
James Harris
===
Subject: Re: Problem with definition of algebraic integers
> 1. First the problematic definition:
> Algebraic integers are defined to be roots of monic
polynomials with
> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 +
17, where
> monic refers to the leading coefficient.
> My assertion is that the over hundred year old definition
excludes
> numbers that have to be included to keep from having
contradiction
> i.e. mathematical inconsistency.
The only problems that you can find with a 
definition is that
there are
no instances of it. This isnÕt a problem, per se, but can be
somewhat
inconvenient. Perhaps you have a problem with one of its
supposed
properties, such as being a ring?
[rest deleted]
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Core error argument objection refuted, short
>IÕve noted a problem in algebraic number theory with the
inclusiveness
>of the definition of algebraic integers as roots of monic
polynomials
>with integer coefficients.
>Various posters have argued that in fact there is no
problem, but
>hereÕs a short refutation of their primary objection.
>First I have
>P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3 f)
>and the factorization
>P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>where the aÕs are given by the following cubic:
>a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
>>If the a_iÕs depend on m, why donÕt you 
write it properly?
>>P(m) = ( a_1(m) x + u f ) ( a_2(m) x + u f ) ( a_3(m) x + u
f )
> ThatÕs a style issue. I see it as a gesture of futility 
and
anguish
> at the reality of this easy refutation.
> Mathematicians are such babies.
ItÕs not a style issue, itÕs a clarity issue. 
If you make your
arguments clear by using precise notation, it becomes easier
to follow
your arguments and it may be easier to see where your
mistakes are.
>My finding that only two of the aÕs have f as a 
factor
without regard
>to the value of m has been vigorously disputed.
>However, consider w_1(m), a factor of a_1 that is a factor
of f, as
>well as a function that varies with m, then it follows that
>a_1 x + uf has w_1(m) as a factor,
>>i.e. w_1(m) | a_1(m) x + u f for all algebraic numbers m
> Mathematicians in trying to deny the reality here have been
hoping on
> some w_1(m) that can give a variable factor of f for a_1.
IÕm just
> following that idea through to the necessary conclusion.
Good. Now if youÕd accept that it exists, you could admit
your error
and we could all move on. It has been shown in raw numbers to
be the case.
>so dividing through by w_1(m) gives
>a_1 x/w_1(m) + uf/w_1(m)
>but then uf/w_1(m) cannot in general be an algebraic integer
as itÕs
>not representable as a polynomial with a finite number of
terms if
>w_1(m) varies with m.
>>I hope you realize that the functions a_i are also not
given by
polynomials
>>in m (i.e., there does not exists a polynomial A_i(M) in
A[M], the
> ThatÕs a rather stupid lie given that I put the polynomial
in this
> post.
> Again it is
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)
> and its roots are a_1, a_2 and a_3.
The aÕs being *roots of a polynomial* in m does not make
*them*
polynomials in m.
> But mathematicians have been persistent in showing a rather
pathetic
> and stupid refusal to accept rather basic mathematics.
> I find them disgusting.
>>polynomial ring in one variable over the algebraic numbers,
such that
>>A_i(m) = A_i(m) for all algebraic numbers m).
>>For the function w_1, well, the only thing youÕve 
mentioned
about it is
>>that w_1(m) | a_1(m) and w_1(m) | f (for all algebraic
numbers m,
>>I think).
>>So how could one expect a_1(m) x / w_1(m) + u f / w_1(m),
or u f / w_1(m)
>>to have any particular form at all? Of course, for u f /
w_1(m)
>>to be an algebraic integer for all algebraic integers m, it
doesnÕt
>>have to be expressible as a polynomial in m over the
algebraic integers.
> The poster is babbling, possibly in shock. Such weakness
from the
> math world is telling. Mathematicians donÕt know how to
really think,
> but have gotten away with faking it.
>>With all you assumptions, it is of course trivial that u f
/ w_1(m)
>>is an algebraic integer (since f is divisible by w_1(m) in
the algebraic
>>integers).
> ItÕs not possible. IÕll put in values to 
help those readers
confused
> by symbols (though algebra is BASED on symbols) by letting
u=2, f=13,
> then you have
> 26/w_1(m)
> and you donÕt have to be a rocket scientist to know that 
no
function
> w_1(m) that actually varies with m can exist such that
26/w_1(m) is an
> algebraic integer for *all* integer m.
Sure there is. w_1(m) = 13 if m>0, w_1(m)=2 if m <= 0. The
actual
w_1(m) will be rather more complicated, and may not have a
nice
representation, but that doesnÕt mean it 
doesnÕt exist.
> ItÕs a show of how broken math society is that Peter van
Rossum would
> dare to make that stupid assertion.
>My guess is that some may be assuming that f is replaceable
by some
>function of m, but in fact, its independent of the value of
m,
>>Everybody understands that, IÕm sure.
>so itÕs
>like 1/(x+1) which is also not representable by a polynomial
if x is
>an algebraic integer not equal to 0 or -2.
>>But I wonder who understands this. I definitely 
donÕt.
>>(By the way, if x *is* an algebraic integer unequal to -1,
then
>>1/(x+1) *is* expressible as a polynomial over the algebraic
integers -
>>a constant polynomial. But you probably mean that there is
no
>>polynomial F(X) over the algebraic integers such that F(x)
= 1/(x+1)
>>for all algebraic integers unequal to -1. I still wonder
what 0 and
>>-2 have to do with it - maybe just a mistake.)
> I was thinking about x being an integer, though in
algebraic integers,
> any x such that x=u_1 - 1, where u_1 is a unit in algebraic
integers
> will work.
>So the objection is refuted by the impossibility of
uf/w_1(m) being an
>algebraic integer, for all algebraic integers m, if w_1(m)
varies with
>m.
>>Can you repeat the definition of algebraic integer again for
>>the newsgroup and tell us how you conclude that u f / w_1(m)
>>is not an algebraic integer?
> Mathematicians are pathetic liars.
> IÕll use u=2, f=13 again, now then, NO function in
algebraic integers
> exists such taht 26/w_1(m) is an algebraic integer for all
integers m,
> if w_1(m) varies with m.
> ItÕs just not possible, but it takes a mathematician to 
lie
about it.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Core error argument objection refuted, short
> IÕve noted a problem in algebraic number theory with the
inclusiveness
> of the definition of algebraic integers as roots of monic
polynomials
> with integer coefficients.
> Various posters have argued that in fact there is no
problem, but
> hereÕs a short refutation of their primary objection.
> First I have
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3 f)
> and the factorization
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> where the aÕs are given by the following cubic:
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
> My finding that only two of the aÕs have f as 
a factor
without regard
> to the value of m has been vigorously disputed.
Perhaps because there is a standing counter-example.
> However, consider w_1(m), a factor of a_1 that is a factor
of f, as
> well as a function that varies with m, then it follows that
> a_1 x + uf has w_1(m) as a factor,
> so dividing through by w_1(m) gives
> a_1 x/w_1(m) + uf/w_1(m)
> but then uf/w_1(m) cannot in general be an algebraic
integer as itÕs
> not representable as a polynomial with a finite number of
terms if
> w_1(m) varies with m.
It would, however, be a *function* which represents different
algebraic
integers for various values of m.
> My guess is that some may be assuming that f is replaceable
by some
> function of m, but in fact,
No. The aÕs are functions of m. f is a constant. Since we
have been
very clear about where the problem is, you are showing low
comprehension
of what people are saying to you.
> its independent of the value of m, so itÕs
> like 1/(x+1) which is also not representable by a
polynomial if x is
> an algebraic integer not equal to 0 or -2.
> So the objection is refuted by the impossibility of
uf/w_1(m) being an
> algebraic integer, for all algebraic integers m, if w_1(m)
varies with
> m.
You have completely failed to understand where the problem
lies. You
have not refuted the counter-example or the
counter-arguments. You have
not even addressed them.
> My hope is that posters who have been so successful in
convincing
> others that my argument is ßawed will post concessions.
Hope springs eternal, does it not?
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Core error argument objection refuted, short
> IÕve noted a problem in algebraic number theory with the
inclusiveness
> of the definition of algebraic integers as roots of monic
polynomials
> with integer coefficients.
> Various posters have argued that in fact there is no
problem, but
> hereÕs a short refutation of their primary objection.
...
> My hope is that posters who have been so successful in
convincing
> others that my argument is ßawed will post concessions.
When do you anticipate that youÕll succeed, James? I imagine
a sort
of steely-eyed General Patton response: As long as it takes,
damn it!
ThatÕs the Army spirit . . .
But what if ten years from now youÕre still presenting them
with an
argument whose truth is so blatantly obvious to you that you
begin
to wonder if they even understand the words youÕre using? 
You
write
number; they read plumber.
Maybe this is just fine with you. Sometimes it seems that you
enjoy
the struggle so much -- your intellect shining out into
unquenchable
darkness -- that deep down you no longer even want victory.
Or maybe you realize that the path youÕve chosen is fraught
with the
possibility of failure, not because of an immovable obstacle
ahead,
but, more insidiously, because it could be like a dream path
which
extends ever onward toward a receding, illusory goal. (But
with so
much drama along the way!)
You believe that the path youÕre on is the only path which
leads to
your goal. But look! Come this way! You protest that itÕs 
not
a
path at all, that it is mere wandering in the woods, and
that, worst
of all, itÕs going almost backwards from your goal! No
promising
vistas. No bright sun. And for GodÕs sake, take off that
Walkman!
You didnÕt even realize, did you? What were you listening 
to?
Magidin Ullrich Overdrive? Now tie this rope around your
waist.
WeÕre going down the cliff. That smoke comes from a village 
I
hope
to reach before nightfall.
But thatÕs a lot of silly talk to an Army man, right? Dig 
the
trench.
Brace the cannon. Call artillery for coordinates.
What will you do in twenty years if youÕre still 
fighting the
same
battle? How long is too long, James?
===
Subject: Re: Core error argument objection refuted, short
>>
>> [.snip.]
>>
>>ItÕs not possible. IÕll put in values to 
help those readers
confused
>>by symbols (though algebra is BASED on symbols) by letting
u=2, f=13,
>>then you have
>>26/w_1(m)
>>and you donÕt have to be a rocket scientist to know that 
no
function
>>w_1(m) that actually varies with m can exist such that
26/w_1(m) is
an
>>algebraic integer for *all* integer m.
>>
>>
>> w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)}
>>
>> IÕm not saying thatÕs the function in 
question (it is
not), but there
>> is a function that actually varies with m such that
26/w_1(m) is an
>> algebraic integer for *all* integer m. So the argument
that no such
>> function can exist is simply bogus.
>Well that example works, so now IÕll say algebraic integer
m, and it
>blows up at m^2+1 = 0.
> So let me see if I have this right:
> (1) You claimed something was impossible under a given set
of
> conditions.
> (2) When I proved that your claim was simply false, you
changed the
> set of conditions, and now argue that what I said was wrong.
You fucking dumbass Arturo Magidin, the ring has ALWAYS been
algebraic
integers, so having m in the ring of algebraic integers
FOLLOWS ANYWAY
you goddamn, fucking, stupid dumbass.
You are so fucking stupid to supposedly be a mathematician
you piss of
shit Arturo Magidin.
You are shit Arturo Magidin. You are a stupid Magidin piece
of shit.
You know what Arturo Magidin? You are a fucking dumbass.
James Harris
===
Subject: Re: Core error argument objection refuted, short
>
> [.snip.]
>
>ItÕs not possible. IÕll put in values to help 
those readers
confused
>by symbols (though algebra is BASED on symbols) by letting
u=2,
f=13,
>then you have
26/w_1(m)
and you donÕt have to be a rocket scientist to know that no
function
>w_1(m) that actually varies with m can exist such that
26/w_1(m) is
an
>algebraic integer for *all* integer m.
>
>
> w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)}
>
> IÕm not saying thatÕs the function in 
question (it is not),
but there
> is a function that actually varies with m such that
26/w_1(m) is an
> algebraic integer for *all* integer m. So the argument that
no such
> function can exist is simply bogus.
>>Well that example works, so now IÕll say algebraic integer
m, and it
>>blows up at m^2+1 = 0.
>> So let me see if I have this right:
>> (1) You claimed something was impossible under a given set
of
>> conditions.
>> (2) When I proved that your claim was simply false, you
changed the
>> set of conditions, and now argue that what I said was
wrong.
>You fucking dumbass Arturo Magidin, the ring has ALWAYS been
algebraic
>integers, so having m in the ring of algebraic integers
FOLLOWS ANYWAY
>you goddamn, fucking, stupid dumbass.
You asked for a function that satisfied certain, 
specific,
explicitly
given properties, claiming such a function was impossible.
I gave a function that satisfied EACH AND EVERY ONE of the
properties
you listed.
Then you changed the properties you wanted. I noted the
change, and
proceeded to give you a function that satisified all the NEW
properties, AND MORE (being defined over all the algebraic
numbers, not
just the algebraic integers; you can restrict it to the
algebraic
integers if you want). But you deleted it. Here it is again:
-- Begin Insert --
HereÕs an example that works for EVERY algebraic NUMBER m:
Step 1. Given an algebraic number m, let f(x) be the unique
monic
polynomial with rational coefficients, irreducible over Q,
which has m
as a root. Write it as:
f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0.
Note that a_0 must be different from 0.
Step 2. Write a_0, a rational number, as a_0 = r/s, with r
and s
integers, r and s relatively prime.
Step 3. If r=1 or -1, let q = 1.
Step 4. If r is not equal to 1 or -1, then let q be the
largest
rational prime that divides r.
Step 5. Let w(m) be a root of the polynomial
x^4 + 13qx^3 + qx + 13.
Then:
(a) w(m) is an algebraic integer.
(b) w(m) divides 13 in the ring of algebraic integers, since
the
product of all the roots is 13, and every root is an algebraic
integer.
(c) w(m) is not constant: it takes different values at
different
integers.
(d) w(m) takes each value a countably infinite number of
times, so
w(m) cannot be given by a polynomial.
-- End Insert --
I gave a variant elsewhere defined for all complex numbers m
as well.
Still think such a function is impossible?
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A manÕs capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A
great
many people are staggered to this extent, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Core error argument objection refuted, short
>> So let me see if I have this right:
>> (1) You claimed something was impossible under a given set
of
>> conditions.
>> (2) When I proved that your claim was simply false, you
changed the
>> set of conditions, and now argue that what I said was
wrong.
> You ******* ******* Arturo Magidin, the ring has ALWAYS
been algebraic
> integers, so having m in the ring of algebraic integers
FOLLOWS ANYWAY
> you *******, *******, stupid *******.
> You are so ******* stupid to supposedly be a mathematician
you **** of
> **** Arturo Magidin.
> You are **** Arturo Magidin. You are a stupid Magidin piece
of ****.
> You know what Arturo Magidin? You are a ******* *******.
> James Harris
Here James once again demonstrates what happens when he sees
that he is
wrong -- he loses control and starts foaming at the mouth.
Here we also
see, once again, what a low-class, filthy-minded, utterly
degenerate
person he is. Beneath all his pretensions of intellect,
logic, and
superiority lies a profound recognition (and abhorrence) of
his own
inferiority; and his reaction when forced to face the truth
about himself
and his depravity is to resort to foul, puerile language as
he rants,
raves, and hurls abuse at all he sees as better than himself.
He is a
perennial dirty-faced, dirty-minded and squalid little boy
throwing mud
at the neatly-dressed people he sees walking by because he
knows he is
not fit for their company.
How ashamed his family must be of him.
--
Wayne Brown (HPCC #1104) | When your tailÕs in a crack, you
improvise
fwbrown@bellsouth.net | if youÕre good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: Core error argument objection refuted, short
> You fucking dumbass Arturo Magidin, the ring has ALWAYS
been algebraic
> integers, so having m in the ring of algebraic integers
FOLLOWS ANYWAY
> you goddamn, fucking, stupid dumbass.
> You are so fucking stupid to supposedly be a mathematician
you piss of
> shit Arturo Magidin.
> You are shit Arturo Magidin. You are a stupid Magidin piece
of shit.
> You know what Arturo Magidin? You are a fucking dumbass.
> James Harris
Are you related to Dar Kabatoff, by any chance?
--
Proposed slogan for JSH: Veni, vidi, vomito.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: Core error argument objection refuted, short
...
> However, consider w_1(m), a factor of a_1 that is a factor
of f, as
> well as a function that varies with m, then it follows that
>
> a_1 x + uf has w_1(m) as a factor,
>
> so dividing through by w_1(m) gives
>
> a_1 x/w_1(m) + uf/w_1(m)
>
> but then uf/w_1(m) cannot in general be an algebraic
integer as itÕs
> not representable as a polynomial with a finite number of
terms if
> w_1(m) varies with m.
>
> A totally off-the-wall, unjustified statement, and, as it
> so happens, incorrect. But for now, if you want to claim
> it is true, the shoe is on your foot: try to prove it.
> IÕve introduced r(m), to handle the result of uf/w_1(m).
> So the poster is requesting that I prove that
> r(m) w_1(m) = uf, does not exist over the ring of algebraic
integers.
> IÕve concluded that using numbers for u and f, as they are
> *independent* of m, helps, so let u=2, f=13.
> Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m),
so you have
> xy=26
> and now IÕll chat further.
> Then the question is, does their exist a multiplicative
inverse in the
> ring of algebraic integers for 26/w_1(m) for *all*
algebraic integers
> m?
> The simple answer is that if w_1(m) varies with m, then it
must vary
> over an infinite number of algebraic integer values as m
varies over
> algebraic integers.
Why? Please prove that.
> But w_1(m) must vary from 0 to infinity if it varies with m.
Why? Please prove that.
> So if you had any doubts about how low mathematicians could
go,
> consider that after all, theyÕre trying to convince that
you can have
> one algebraic integer function defined by the multiplicative
inverse
> of another algebraic integer function over all algebraic
integer m,
> like with f=7, 14/w_1(m) = r(m).
> ItÕs like saying that you can have xy=2, where x and y are
integers,
> or algebraic integers, where x varies over the ring, and y
remains in
> it, though of course, I can just show that for x=5 that
doesnÕt work.
Not at all. The range of w_1(m) may be severely limited,
while the
range of x is not limited. It is similar to saying that you
can have
xy = 2, where x ranges over the algebraic integer factors of
2, and so y
also ranges over the algebraic integer factors of 2. 5 is now
not
a counterexample because it is not an algebraic integer
factor of 2.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Core error argument objection refuted, short
> ...
> > However, consider w_1(m), a factor of a_1 that is a
factor of f,
as
> > well as a function that varies with m, then it follows
that
> >
> > a_1 x + uf has w_1(m) as a factor,
> >
> > so dividing through by w_1(m) gives
> >
> > a_1 x/w_1(m) + uf/w_1(m)
> >
> > but then uf/w_1(m) cannot in general be an algebraic
integer as
itÕs
> > not representable as a polynomial with a finite number of
terms if
> > w_1(m) varies with m.
> >
> > A totally off-the-wall, unjustified statement, and, as it
> > so happens, incorrect. But for now, if you want to claim
> > it is true, the shoe is on your foot: try to prove it.
> >
> > IÕve introduced r(m), to handle the result of uf/w_1(m).
> >
> > So the poster is requesting that I prove that
> >
> > r(m) w_1(m) = uf, does not exist over the ring of
algebraic integers.
> >
> > IÕve concluded that using numbers for u and f, as they 
are
> > *independent* of m, helps, so let u=2, f=13.
> >
> > Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m),
so you have
> >
> > xy=26
> >
> > and now IÕll chat further.
> >
> > Then the question is, does their exist a multiplicative
inverse in the
> > ring of algebraic integers for 26/w_1(m) for *all*
algebraic integers
> > m?
> >
> > The simple answer is that if w_1(m) varies with m, then
it must vary
> > over an infinite number of algebraic integer values as m
varies over
> > algebraic integers.
> Why? Please prove that.
It has to do with continuity and slope. If you could have
w_1(m) and
w_1(mÕ) equal when m does not equal mÕ then at 
that point
youÕd have
infinite slope or a discontinuity.
IÕm not interested in explaining basic mathematics but at
least
hopefully I can give other readers some sense of the
frustration IÕve
had to handle dealing with posters who are so mathematically
ignorant,
while fanatically replying to my posts negatively.
> > But w_1(m) must vary from 0 to infinity if it varies with
m.
> Why? Please prove that.
It turns out that you need the the absolute value, like
r(m)r*(m), and
with it itÕs possible to show that for an algebraic integer
function
that varies as m varies--a continuous function--as m varies
over all
of algebraic integer r(m)r*(m) must vary from 0 to positive or
negative infinity.
> > So if you had any doubts about how low mathematicians
could go,
> > consider that after all, theyÕre trying to convince that
you can have
> > one algebraic integer function defined by the
multiplicative inverse
> > of another algebraic integer function over all algebraic
integer m,
> > like with f=7, 14/w_1(m) = r(m).
> >
> > ItÕs like saying that you can have xy=2, where x and y
are integers,
> > or algebraic integers, where x varies over the ring, and
y remains in
> > it, though of course, I can just show that for x=5 that
doesnÕt work.
> Not at all. The range of w_1(m) may be severely limited,
while the
> range of x is not limited. It is similar to saying that you
can have
> xy = 2, where x ranges over the algebraic integer factors
of 2, and so y
> also ranges over the algebraic integer factors of 2. 5 is
now not
> a counterexample because it is not an algebraic integer
factor of 2.
Yeah you can *say* just about anything, but mathematically
that
statement is bullshit, and for me to have made the
discoveries IÕve
made and be stuck because a lot of posters can get away with
bullshit
is just pissing me off.
James Harris
===
Subject: Re: Core error argument objection refuted, short
...
> > The simple answer is that if w_1(m) varies with m, then
it must
vary
> > over an infinite number of algebraic integer values as m
varies
over
> > algebraic integers.
>
> Why? Please prove that.
> It has to do with continuity and slope. If you could have
w_1(m) and
> w_1(mÕ) equal when m does not equal mÕ then 
at that point
youÕd have
> infinite slope or a discontinuity.
Ah, you assume a continuous function. Why?
> > But w_1(m) must vary from 0 to infinity if it varies with
m.
>
> Why? Please prove that.
> It turns out that you need the the absolute value, like
r(m)r*(m), and
> with it itÕs possible to show that for an algebraic 
integer
function
> that varies as m varies--a continuous function--as m varies
over all
> of algebraic integer r(m)r*(m) must vary from 0 to positive
or
> negative infinity.
Assuming a continuous function again. Why?
> > ItÕs like saying that you can have xy=2, where x and y 
are
integers,
> > or algebraic integers, where x varies over the ring, and
y remains
in
> > it, though of course, I can just show that for x=5 that
doesnÕt
work.
>
> Not at all. The range of w_1(m) may be severely limited,
while the
> range of x is not limited. It is similar to saying that you
can have
> xy = 2, where x ranges over the algebraic integer factors
of 2, and so
y
> also ranges over the algebraic integer factors of 2. 5 is
now not
> a counterexample because it is not an algebraic integer
factor of 2.
> Yeah you can *say* just about anything, but mathematically
that
> statement is bullshit, and for me to have made the
discoveries IÕve
> made and be stuck because a lot of posters can get away
with bullshit
> is just pissing me off.
You are producing the bullshit. Pray produce a proof that
w_1(m) is a
continuous function and we can talk further.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Core error argument objection refuted, short
> ...
> > However, consider w_1(m), a factor of a_1 that is a
factor of f,
as
> > well as a function that varies with m, then it follows
that
> >
> > a_1 x + uf has w_1(m) as a factor,
> >
> > so dividing through by w_1(m) gives
> >
> > a_1 x/w_1(m) + uf/w_1(m)
> >
> > but then uf/w_1(m) cannot in general be an algebraic
integer as
itÕs
> > not representable as a polynomial with a finite number of
terms
if
> > w_1(m) varies with m.
> >
> > A totally off-the-wall, unjustified statement, and, as it
> > so happens, incorrect. But for now, if you want to claim
> > it is true, the shoe is on your foot: try to prove it.
> >
> > IÕve introduced r(m), to handle the result of uf/w_1(m).
> >
> > So the poster is requesting that I prove that
> >
> > r(m) w_1(m) = uf, does not exist over the ring of
algebraic
integers.
> >
> > IÕve concluded that using numbers for u and f, as they 
are
> > *independent* of m, helps, so let u=2, f=13.
> >
> > Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m),
so you
have
> >
> > xy=26
> >
> > and now IÕll chat further.
> >
> > Then the question is, does their exist a multiplicative
inverse in
the
> > ring of algebraic integers for 26/w_1(m) for *all*
algebraic
integers
> > m?
> >
> > The simple answer is that if w_1(m) varies with m, then
it must vary
> > over an infinite number of algebraic integer values as m
varies over
> > algebraic integers.
>
> Why? Please prove that.
> It has to do with continuity and slope. If you could have
w_1(m) and
> w_1(mÕ) equal when m does not equal mÕ then 
at that point
youÕd have
> infinite slope or a discontinuity.
OOPS! What I said was STUPID!!!
WhatÕs interesting about that error is that you can see
replies to it
in this thread.
I want you all to *focus* on the replies. Read them carefully.
Oh yeah, so how do you prove that a varying function in
algebraic
integers has to have an infinite number of results?
Anybody? Anybody?
James Harris
===
Subject: Re: Core error argument objection refuted, short
...
> Why? Please prove that.
>
> It has to do with continuity and slope. If you could have
w_1(m) and
> w_1(mÕ) equal when m does not equal mÕ then 
at that point
youÕd have
> infinite slope or a discontinuity.
> OOPS! What I said was STUPID!!!
Yes.
> Oh yeah, so how do you prove that a varying function in
algebraic
> integers has to have an infinite number of results?
> Anybody? Anybody?
The simple answer is: you canÕt. For instance the function
(assume x an
algebraic integer):
f(x) = 2 when x/2 is an algebraic integer,
f(x) = 1 otherwise,
is a perfectly well-defined varying function in the function
that has
only two possible results.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Core error argument objection refuted, short
> ...
> > Why? Please prove that.
> >
> > It has to do with continuity and slope. If you could have
w_1(m)
and
> > w_1(mÕ) equal when m does not equal mÕ 
then at that point
youÕd have
> > infinite slope or a discontinuity.
> >
> > OOPS! What I said was STUPID!!!
> Yes.
> > Oh yeah, so how do you prove that a varying function in
algebraic
> > integers has to have an infinite number of results?
> >
> > Anybody? Anybody?
> The simple answer is: you canÕt. For instance the function
(assume x an
> algebraic integer):
> f(x) = 2 when x/2 is an algebraic integer,
> f(x) = 1 otherwise,
> is a perfectly well-defined varying function in the function
that has
> only two possible results.
Wow that poster is astoundingly stupid as if youÕre *in the
ring of
algebraic integers* how can you have x/2 NOT be an algebraic
integer?
Trick question: Remember, the / operator is NOT defined in
general
in the ring, and by convention you can only use it when a
given
numerator has the denominator as a factor.
Therefore, to use x/2 it must be that x is even, or the
condition is
undefined in the ring.
James Harris
===
Subject: Re: Core error argument objection refuted, short
>> ...
>> > Why? Please prove that.
>> >
>> > It has to do with continuity and slope. If you could
have w_1(m)
and
>> > w_1(mÕ) equal when m does not equal mÕ 
then at that
point youÕd
have
>> > infinite slope or a discontinuity.
>> >
>> > OOPS! What I said was STUPID!!!
>>
>> Yes.
>>
>> > Oh yeah, so how do you prove that a varying function in
algebraic
>> > integers has to have an infinite number of results?
>> >
>> > Anybody? Anybody?
>> The simple answer is: you canÕt. For instance the 
function
(assume x an
>> algebraic integer):
>> f(x) = 2 when x/2 is an algebraic integer,
>> f(x) = 1 otherwise,
>> is a perfectly well-defined varying function in the
function that has
>> only two possible results.
>Wow that poster is astoundingly stupid as if youÕre *in the
ring of
>algebraic integers* how can you have x/2 NOT be an algebraic
integer?
If x=2, x is an algebraic integer but x/2 is not.
If x=5, x is an algebraic integer, but x/2 is not.
If x = sqrt(2), x is an algebraic integer, but x/2 is not.
>Trick question: Remember, the / operator is NOT defined in
general
>in the ring, and by convention you can only use it when a
given
>numerator has the denominator as a factor.
>Therefore, to use x/2 it must be that x is even, or the
condition is
>undefined in the ring.
Unusually idiotic even for you. But if you insist...
Let x/2 represent the division of x by 2 in the complex
numbers. For
every algebraic integer x, let
f(x) = 2 when x/2 is an algebraic integer,
f(x) = 1 otherwise.
I hope even you realize that you can take an x from the
algebraic
integers, calculate the number x/2, and check whether it too
is an
algebraic integer.
- Randy
===
Subject: Re: Core error argument objection refuted, short
> ...
> > Why? Please prove that.
> >
> > It has to do with continuity and slope. If you could have
w_1(m)
and
> > w_1(mÕ) equal when m does not equal mÕ 
then at that point
youÕd
have
> > infinite slope or a discontinuity.
> >
> > OOPS! What I said was STUPID!!!
>
> Yes.
>
> > Oh yeah, so how do you prove that a varying function in
algebraic
> > integers has to have an infinite number of results?
> >
> > Anybody? Anybody?
>
> The simple answer is: you canÕt. For instance the function
(assume x
an
> algebraic integer):
> f(x) = 2 when x/2 is an algebraic integer,
> f(x) = 1 otherwise,
> is a perfectly well-defined varying function in the function
that has
> only two possible results.
> Wow that poster is astoundingly stupid as if youÕre *in 
the
ring of
> algebraic integers* how can you have x/2 NOT be an
algebraic integer?
That poster is astoundingly stupid as if youÕre *in the ring
of
algebraic integers* when the only thing I said is that *x is
an
algebraic integer*. And it is easily shown that f is a
function
from the algebraic integers to the algebraic integers (which
you
call as being in the algebraic integers).
> Therefore, to use x/2 it must be that x is even, or the
condition is
> undefined in the ring.
Nonsense.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Core error argument objection refuted, short
are you saying that the algebraic integers are continuous
on hte real line, or just incredibly dense?... or,
just taht there are an infinity of them?
> I want you all to *focus* on the replies. Read them
carefully.
> Oh yeah, so how do you prove that a varying function in
algebraic
> integers has to have an infinite number of results?
--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...
La Troi Phases dÕExploitation de la Protocols des Grises de
Kyoto:
(FOSSILISATION [McCainanites?] (TM/sic))/
BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm.
Http://www.tarpley.net/bushb.htm (content partiale, below):
17 -- LÕATTEMPTER de COUP DÕETAT, 3/30/81
===
Subject: Re: Core error argument objection refuted, short
>> ...
>> > However, consider w_1(m), a factor of a_1 that is a
factor of f,
as
>> > well as a function that varies with m, then it follows
that
>> >
>> > a_1 x + uf has w_1(m) as a factor,
>> >
>> > so dividing through by w_1(m) gives
>> >
>> > a_1 x/w_1(m) + uf/w_1(m)
>> >
>> > but then uf/w_1(m) cannot in general be an algebraic
integer as
itÕs
>> > not representable as a polynomial with a finite number of
terms
if
>> > w_1(m) varies with m.
>> >
>> > A totally off-the-wall, unjustified statement, and, as it
>> > so happens, incorrect. But for now, if you want to claim
>> > it is true, the shoe is on your foot: try to prove it.
>> >
>> > IÕve introduced r(m), to handle the result of 
uf/w_1(m).
>> >
>> > So the poster is requesting that I prove that
>> >
>> > r(m) w_1(m) = uf, does not exist over the ring of
algebraic integers.
>> >
>> > IÕve concluded that using numbers for u and f, as they
are
>> > *independent* of m, helps, so let u=2, f=13.
>> >
>> > Then you have r(m) w_1(m) = 26, and let x=r(m),
y=w_1(m), so you have
>> >
>> > xy=26
>> >
>> > and now IÕll chat further.
>> >
>> > Then the question is, does their exist a multiplicative
inverse in
the
>> > ring of algebraic integers for 26/w_1(m) for *all*
algebraic integers
>> > m?
>> >
>> > The simple answer is that if w_1(m) varies with m, then
it must vary
>> > over an infinite number of algebraic integer values as m
varies over
>> > algebraic integers.
>> Why? Please prove that.
>It has to do with continuity and slope. If you could have
w_1(m) and
>w_1(mÕ) equal when m does not equal mÕ then 
at that point
youÕd have
>infinite slope or a discontinuity.
Whee! In another post you just said
The function has to be continuous given the other equations
which not
surprisingly the poster deleted. These people are dumb as
rocks.
and I considered asking whether you could _define_ the word
continuous. Decided not to bother, because continuity is one
of those things that a person _could_ have a fairly reasonable
understanding of even if he were unable to give the precise
definition.
But here you show you have _no_ clue regarding continuity
and slope, which is funnier than your cluelessness regarding
algebraic number theory, because continuity and slope
were things you were supposed to learn about when you
got that famous degree in physics.
Hint: If you could have w_1(m) and w_1(mÕ) equal when m does
not equal mÕ then at that point youÕd have 
infinite slope or
a discontinuity is funny. Like say f(m) = 42; f is constant.
It follows from what you say here that a _constant_ function
has infinite slope or a discontinuity. Wow.
>IÕm not interested in explaining basic mathematics but at
least
>hopefully I can give other readers some sense of the
frustration IÕve
>had to handle dealing with posters who are so mathematically
ignorant,
>while fanatically replying to my posts negatively.
Yeah. So mathematically ignorant that they donÕt realize 
that
constant functions are discontinuous (or that integers are
irrational, that sqrt(i) is not a complex number, etc... 
itÕs
really
remarkable how ignorant we all are. <guffaw>)
>> > But w_1(m) must vary from 0 to infinity if it varies with
m.
>> Why? Please prove that.
>It turns out that you need the the absolute value, like
r(m)r*(m), and
>with it itÕs possible to show that for an algebraic integer
function
>that varies as m varies--a continuous function--as m varies
over all
>of algebraic integer r(m)r*(m) must vary from 0 to positive
or
>negative infinity.
>> > So if you had any doubts about how low mathematicians
could go,
>> > consider that after all, theyÕre trying to convince 
that
you can have
>> > one algebraic integer function defined by the
multiplicative inverse
>> > of another algebraic integer function over all algebraic
integer m,
>> > like with f=7, 14/w_1(m) = r(m).
>> >
>> > ItÕs like saying that you can have xy=2, where x and y
are integers,
>> > or algebraic integers, where x varies over the ring, and
y remains in
>> > it, though of course, I can just show that for x=5 that
doesnÕt work.
>> Not at all. The range of w_1(m) may be severely limited,
while the
>> range of x is not limited. It is similar to saying that
you can have
>> xy = 2, where x ranges over the algebraic integer factors
of 2, and so y
>> also ranges over the algebraic integer factors of 2. 5 is
now not
>> a counterexample because it is not an algebraic integer
factor of 2.
>Yeah you can *say* just about anything, but mathematically
that
>statement is bullshit, and for me to have made the
discoveries IÕve
>made and be stuck because a lot of posters can get away with
bullshit
>is just pissing me off.
>James Harris
************************
David C. Ullrich
===
Subject: Re: Core error argument objection refuted, short
> IÕm not interested in explaining basic mathematics but at
least
> hopefully I can give other readers some sense of the
frustration IÕve
> had to handle dealing with posters who are so
mathematically ignorant,
> while fanatically replying to my posts negatively.
In other words, you have not the slightest idea what you are
talking
about.
===
Subject: Re: Core error argument objection refuted, short
> IÕve noted a problem in algebraic number theory with the
inclusiveness
> of the definition of algebraic integers as roots of monic
polynomials
> with integer coefficients.
>
> Various posters have argued that in fact there is no
problem, but
> hereÕs a short refutation of their primary objection.
>
> First I have
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3 f)
>
> and the factorization
>
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> where the aÕs are given by the following cubic:
>
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
>
> My finding that only two of the aÕs have f as 
a factor
without regard
> to the value of m has been vigorously disputed.
>
>
> ItÕs incorrect. See below.
> Mathematicians, you see, have other priorities than actual
validity of
> mathematics, as they have *social* issue. See below.
Instantly, we start off with the character attack. Why?
> However, consider w_1(m), a factor of a_1 that is a factor
of f, as
> well as a function that varies with m, then it follows that
>
> a_1 x + uf has w_1(m) as a factor,
>
> so dividing through by w_1(m) gives
>
> a_1 x/w_1(m) + uf/w_1(m)
>
> but then uf/w_1(m) cannot in general be an algebraic
integer as itÕs
> not representable as a polynomial with a finite number of
terms if
> w_1(m) varies with m.
>
>
> A totally off-the-wall, unjustified statement, and, as it
> so happens, incorrect. But for now, if you want to claim
> it is true, the shoe is on your foot: try to prove it.
> IÕve introduced r(m), to handle the result of uf/w_1(m).
> So the poster is requesting that I prove that
> r(m) w_1(m) = uf, does not exist over the ring of algebraic
integers.
> IÕve concluded that using numbers for u and f, as they are
> *independent* of m, helps, so let u=2, f=13.
> Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m),
so you have
> xy=26
And where exactly is your proof that this cannot happen?
This is addressed in another thread. It is assuredly very
possible to have functions r(m) and w_1(m) taking values in
the algebraic integers, such that their product is constant.
> and now IÕll chat further.
> Then the question is, does their exist a multiplicative
inverse in the
> ring of algebraic integers for 26/w_1(m) for *all*
algebraic integers
> m?
You donÕt need a multiplicative inverse. For example,
26 = 26^{1/3} * 26^{2/3}.
Neither of these is the multiplicative inverse of the
other. There are of course infinitely many such ways to
factor a nonunit algebraic integer.
> The simple answer is that if w_1(m) varies with m, then it
must vary
> over an infinite number of algebraic integer values as m
varies over
> algebraic integers.
> But w_1(m) must vary from 0 to infinity if it varies with m.
No! - actually its range is quite limited. See my post of Oct
18 in
the other thread, Finishing argument - core error proven.
> Replies IÕve seen have shied away from trying something
like w_1(m) =
> m+1 because most readers can immediately realize that
26/(m+1) or
> anything like it, canÕt be an algebraic integer for all m.
Of course.
> Instead IÕve seen examples like 26^{1/m} or more trying,
26^{1/m^2+1},
> but notice that because m^2+1 can equal 0, *in the ring of
algebraic
> integers*, you can get 26^{1/0} for the mÕs that are the
roots of
> m^2+1.
Your own argument which originated this discussion assumes m
is
an ordinary integer. There is no reason to consider more
general
m. Moreover, Arturo has given w_1(m) which are well-defined
everywhere and fit the requirements.
> My guess is that some may be assuming that f is replaceable
by some
> function of m, but in fact, its independent of the value of
m,
>
> No, we are not assuming f is replaceable by some function
of m.
> We assume f itself is *constant* with respect to m.
However, the
factorization
> of a polynomial P(m) whose coefficients are functions of m,
hence
> dependent on m, is also in general dependent on m. We do
*not* assume
> that the corresponding factorization of f is constant with
respect
> to m.
> So if you had any doubts about how low mathematicians could
go,
> consider that after all, theyÕre trying to convince that
you can have
> one algebraic integer function defined by the multiplicative
inverse
> of another algebraic integer function over all algebraic
integer m,
> like with f=7, 14/w_1(m) = r(m).
Inverses are not an issue. If a*b = 14, there is no
implication
that b is the inverse of a. That only happens when a*b = 1.
> ItÕs like saying that you can have xy=2, where x and y are
integers,
> or algebraic integers, where x varies over the ring, and y
remains in
> it, though of course, I can just show that for x=5 that
doesnÕt work.
There is no reason to worry about x = 5, unless you somehow
know that w_1(m) must take on the value 5 when the product is
2.
You must be thinking that for fixed u*f, w_1(m) must take on
very large values when m gets large, i.e., that it is an
*unbounded*
function, like, say, w_1(m) = 3*m + 4, or w_1(m) = m^{.3}, or
some
such. That is not how it behaves at all. See my post of Oct
18 in
the thread Finishing argument - core error proven.
> Hmmm...thatÕs why posters have tried to pick w_1(m) using
exponential
> functions.
> Fascinating as theyÕre showing how much they understand.
> so itÕs
> like 1/(x+1) which is also not representable by a
polynomial if x is
> an algebraic integer not equal to 0 or -2.
>
> So the objection is refuted by the impossibility of
uf/w_1(m) being
an
> algebraic integer,
>
>
> Also incorrect and false. Remember what you said above:
>
> ... consider w_1(m), a factor of a_1 that is a factor of
f...
>
> If w_1(m) is a factor of f, that can only mean f/w_1(m) is
> an algebraic integer, which of course implies that uf/w_1(m)
> is an algebraic integer: this is your *assumption* here.
> The assumption is itself a contradiction, in that it
requires the
> ability to define one algebraic integer function by the
multiplicative
> inverse of another algebraic integer function.
The point here was: the assumption was YOUR assumption. And
again, there is no issue involving inverses.
> Basic algebra.
> for all algebraic integers m, if w_1(m) varies with
> m.
>
> My hope is that posters who have been so successful in
convincing
> others that my argument is ßawed will post concessions.
>
>
>
> Absolutely not!
>
> You are inching closer to the truth in this. You are seeing
> how this can work. You are correct that w_1(m) is a factor
of
> f that depends on m. As noted above this implies that
> uf/w_1(m) is an A.I., and a1(m)/w_1(m) is also. I think you
> are beginning to see how this can happen.
> Readers should note that attempt to push the impossible,
one algebraic
> integer function defined by the multiplicative inverse of
another
> algebraic integer function.
Readers should note tiresome repetition of this theme. As
was noted in the other thread (Finishing ... ), for the
crucial cases of interest, w_1(m) = f^{2/3}. This never
causes problems
of the kind you are worrying about here, because you are
assuming
w_1(m)*r(m) = u*f.
> My guess is that seeing something like uf/w_1(m) may
confuse some who
> might believe that u and f are functions of m,
Nonsense. We all accept that u and f are constant with
respect to m.
> so I like to show their
> independence by tossing in values, like u=2, f=13, so you
have
> 26/w_1(m).
> I note a couple of other things. First, the polynomial in a
> that you mention above:
>
> [#] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
>
> Your claim is that two roots of this polynomial have a
factor
> that is f. This means that if r is one of those roots, then
> r = f*c, where c is an algebraic integer. This implies that
>
> f^3*c^3 + 3(-1 +m*f^2)*f^2*c^2 - f^2*(m^3*f^4 - 3*m*f^2 +
3*m) = 0,
>
> or, factoring out f^2,
>
> f*c^3 + 3*(-1 + m*f^2)*c^2 - (m^3*f^4 - 3*m*f^2 + 3*m) = 0.
>
>
> The polynomial in c on the left is non-monic and primitive
> (if f is not a multiple of 3). If it is irreducible, then
> c cannot be an algebraic integer. When f = 5 and m = 1, this
> equation is
>
> 5*c^3 + 72*c^2 - 553 = 0,
>
> and the polynomial in c is easily shown to be irreducible.
> Therefore c cannot be an algebraic integer. This contradicts
> your central claim: NONE of the roots of [#] can have a
> factor that is 5 = f.
>
How come you never comment on this? Do you not understand it?
Do you not see why it is relevant and important? DonÕt you
care
that it destroys your claims? Or do you just not have any
refutation?
>
> Second note: It may be worthwhile to see how irreducibility
> is inextricably tied to factorization of roots. Assume that
>
>
> Q(m) = x^2 + m*x + 30,
>
> where m is an integer.
>
> The constant term is divisible by 5 (also by 2 and 3, but I
> will focus on 5 here).
>
> For certain values of m, this polynomial is reducible, and
the
> corresponding roots are not both divisible by 5:
>
> m = 11: r1 = 5, r2 = 6
>
> m = 13: r1 = 10, r2 = 3
>
> m = 17: r1 = 15, r2 = 2
>
> m = 31: r1 = 30, r2 = 1.
>
> In all these examples, obviously one root is *divisible* by
5
> and the other is *coprime* to 5. And of course in all these
> examples, the polynomial is reducible. This is parallel to
> HarrisÕs cubic when m = 0: two of the roots are divisible
> by f and one is coprime to f. In this particular case his
polynomial
> [#] is reducible. In general it is not.
>
> Now look at an other example. Say, m = 14. The
> polynomial is irreducible, because the discriminant
>
> D^2 = m^ - 4*30 = 76,
>
> which is not a perfect square. The roots of the
> polynomial are:
>
>
> r1 = (-14 + sqrt(76))/2 = -7 + sqrt(19) and
>
> r2 = -7 - sqrt(19).
>
> Clearly both r1 and r2 are algebraic integers.
>
> Assume that r1 is a multiple of 5: r1 = 5*s1. Then
>
> 25*s1^2 + 70*s1 + 30 = 0.
>
> Factor out 5:
>
> [*] 5*s1^2 + 14*s1 + 6 = 0.
>
> This happens to have discriminant D^2 = 76. [This is
> not a coincidence!]. Thus the polynomial in [*] is
> *also* irreducible. Therefore s1 cannot be an algebraic
> integer.
>
> Therefore r1 cannot be divisible by 5.
>
> The same can similarly be shown for r2.
>
> However, it is now easy to show that both r1 and r2 must
both
> be *non-coprime* to 5. For, suppose r1 is coprime to 5.
> The fact that
>
> r1*r2 = 5*6
>
> would then imply that r2 is DIVISIBLE by 5, which
> contradicts the result above that both r1 and r2 are
> NOT divisible by 5.
>
> Conclusion: for m = 14, the polynomial is irreducible,
> and both roots have a nonunit algebraic integer factor
> in common with 5.
>
> This is true more generally. There is nothing special
> about 14. Try, for example, m = 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 12,
> 14, 15, 16, 18, ... MOST values of m yield an irreducible
> polynomial, and both of the roots will both share algebraic
> integer factors in common with 5.
>
> In general it is hard to write down exactly what these
> factors are, even in the quadratic case. The important
> thing to know about them here is that, as suggested above
> by the w_1(m) notation, *they will be dependent on m*.
> WhatÕs fascinating here is how casually posters will just
start
> talking, which I think actually usually works with a lot of
readers
> who donÕt even bother to read carefully.
I see no sign here that you read or understood anything in the
preceding section, or saw its relevance to your beliefs about
factorization. Ironically, I think you yourself must be one of
the readers who donÕt even bother to read carefully. Right?
> Just remember that the base position here is that in the
ring of
> algebraic integers you can have one function defined by the
> multiplicative inverse of another function.
> Remember xy=26.
Also remember that 26 / x is *not* the inverse of x.
Otherwise, 13 would be the inverse of 2 !
Nora B.
> James Harris
===
Subject: Re: Core error argument objection refuted, short
> Mathematicians, you see, have other priorities than actual
validity of
> mathematics, as they have *social* issue. See below.
>
> Instantly, we start off with the character attack. Why?
Insecurity and lack of any better arguments.
===
Subject: Re: Core error argument objection refuted, short
In sci.math, Nora Baron
<norabaron@hotmail.com> IÕve noted a problem in algebraic
number theory with the
inclusiveness
>> of the definition of algebraic integers as roots of monic
polynomials
>> with integer coefficients.
>>
>> Various posters have argued that in fact there is no
problem, but
>> hereÕs a short refutation of their primary objection.
>>
>> First I have
>>
>> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x
u^2 + u^3
f)
>>
>> and the factorization
>>
>> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>>
>> where the aÕs are given by the following cubic:
>>
>> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
>>
>> My finding that only two of the aÕs have f as 
a factor
without
regard
>> to the value of m has been vigorously disputed.
>>
>>
>> ItÕs incorrect. See below.
>> Mathematicians, you see, have other priorities than actual
validity of
>> mathematics, as they have *social* issue. See below.
> Instantly, we start off with the character attack. Why?
IsnÕt it obvious? He canÕt attack the bits; 
theyÕre too
small. :-)
He canÕt attack the words; theyÕre too big.
So he attacks the chars in the middle. :-) But he may find
that weÕll take a byte out of him in the end... :-)
(And yes, some mathematicians -- and computer engineers --
have
priorities such as making very bad puns. :-) )
[rest snipped]
--
#191, ewill3@earthlink.net
ItÕs still legal to go .sigless.
===
Subject: Re: Core error argument objection refuted, short
> IÕve noted a problem in algebraic number theory with the
inclusiveness
> of the definition of algebraic integers as roots of monic
polynomials
> with integer coefficients.
>
> Various posters have argued that in fact there is no
problem, but
> hereÕs a short refutation of their primary objection.
>
> First I have
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3
f)
>
> and the factorization
>
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> where the aÕs are given by the following cubic:
>
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
>
> My finding that only two of the aÕs have f as 
a factor
without
regard
> to the value of m has been vigorously disputed.
>
>
> ItÕs incorrect. See below.
>
> Mathematicians, you see, have other priorities than actual
validity of
> mathematics, as they have *social* issue. See below.
>
> Instantly, we start off with the character attack. Why?
You lack character and are a base liar.
YouÕre a despicable human being.
YouÕre disgusting trash.
> However, consider w_1(m), a factor of a_1 that is a factor
of f, as
> well as a function that varies with m, then it follows that
>
> a_1 x + uf has w_1(m) as a factor,
>
> so dividing through by w_1(m) gives
>
> a_1 x/w_1(m) + uf/w_1(m)
>
> but then uf/w_1(m) cannot in general be an algebraic
integer as
itÕs
> not representable as a polynomial with a finite number of
terms if
> w_1(m) varies with m.
>
>
> A totally off-the-wall, unjustified statement, and, as it
> so happens, incorrect. But for now, if you want to claim
> it is true, the shoe is on your foot: try to prove it.
>
> IÕve introduced r(m), to handle the result of uf/w_1(m).
>
> So the poster is requesting that I prove that
>
> r(m) w_1(m) = uf, does not exist over the ring of algebraic
integers.
>
> IÕve concluded that using numbers for u and f, as they are
> *independent* of m, helps, so let u=2, f=13.
>
> Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m),
so you have
>
> xy=26
>
> And where exactly is your proof that this cannot happen?
You are rather stupid when it comes to math you disgusting
piece of
trash.
You canÕt have xy=26 in the ring of algebraic integers with 
x
varying
over algebraic integers *and* y an algebraic integer for all
x.
YouÕre stupid, and despicable.
YouÕre trash Nora Baron.
James Harris
===
Subject: Re: Core error argument objection refuted, short
> IÕve noted a problem in algebraic number theory with the
inclusiveness
> of the definition of algebraic integers as roots of monic
polynomials
> with integer coefficients.
>
> Various posters have argued that in fact there is no
problem, but
> hereÕs a short refutation of their primary objection.
>
> First I have
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2
+ u^3
f)
>
> and the factorization
>
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> where the aÕs are given by the following cubic:
>
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).
>
> My finding that only two of the aÕs have f as 
a factor
without
regard
> to the value of m has been vigorously disputed.
>
>
> ItÕs incorrect. See below.
>
> Mathematicians, you see, have other priorities than actual
validity
of
> mathematics, as they have *social* issue. See below.
>
>
> Instantly, we start off with the character attack. Why?
> You lack character and are a base liar.
> YouÕre a despicable human being.
> YouÕre disgusting trash.
Harris starts with the polynomial
[1] P(m) = f^2 * [(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
- 3*(-1 + m*f^2)*u^2*x + u*3*f]
and assumes a factorization in the form
[2] P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
where a1, a2, and a3 are algebraic integers and functions of
m. Harris claims that two of the aÕs are divisible by f and
the third one is relatively prime to f. I and others here say
that, for most values of m, this is incorrect; that each of
the aÕs has a factor in common with f. This is expressed by
saying
f^2 = w1 * w2 * w3,
where w1, w2, and w3 are algebraic integer factors of f^2,
and w1, w2, and w3 divide a1, a2, and a3 respectively.
Note that w1, w2, and w3 are dependent on m:
w1 = w1(m), w2 = w2(m), and w3 = w3(m).
They are also dependent on u and f, but we will assume those
are fixed values in this discussion.
Yesterday in another thread I stated that, for most values of
m, w1, w2, and w3 are equal to f^{2/3}. I thought I had a
proof
of this. It may be true, and everything seems to fit together
if it is true, but the proof has eluded me.
Here is what I was thinking. Let Q(m) = P(m) / f^2, i.e.,
[3] Q(m) = (m^3*f^4 -3*m^2*f^2 + 3*m)*x^3 - 3*(-1 +
m*f^2)*u^2*x + u*3*f]
Paralleling Harris, I assumed that Q(m) could be factored in
the form
[4] Q(m) = (b1*x + u*f^{1/3})*(b2*x + u*f{1/3})*(b3*x +
u*f^{1/3}).
But this is the part I cannot prove. A more general
factorization,
[5] Q(m) = (b1*x + c1)*(b2*x + c2)*(b3*x + c3) can be assumed,
where all of b1, b2, b3, c1, c2, c3 are algebraic integers.
But
I have not found a way to justify saying that each of the 
cÕs
is equal to u*f^{1/3}. If it is correct, it leads to the
result
noted above, that w1, w2 and w3 are all equal to f^{2/3} for
most values of m [essentially whenever a related cubic
polynomial,
a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m), is
irreducible].
So I may have been wrong about the factorization of Q(m). I
would
like for the factorization [4] to be true, but maybe one has
to
assume the more general factorization [5].
The point I am raising here however is, similarly how can one
justify HarrisÕs factorization [2] ? Possibly in that case
also the
most we can assume is
[6] P(m) = (a1*x + d1)*(a2*x + d2)*(a3*x + d3),
where a1, a2, a3 and d1, d2, d3 are algebraic integers. That
is,
again, how can Harris assume that each of the dÕs is equal 
to
u*f ?
We have been arguing about the factorization [2] for months.
We
have proofs that, if this factorization *is* assumed, then
HarrisÕs
claim that two of the aÕs are divisible by f is false in
general
(again depending on irreducibility of the related polynomial).
Those proofs are valid when factorization [2] is assumed.
However those proofs do not work if the more general
factorizations
[6] or [5] are assumed. In that case it can be shown that it
is
entirely possible that two of the aÕs are divisible by f.
Whether
this is sufficient for HarrisÕs purposes I 
donÕt know.
So the real questions here are: can one assume factorizations
of
the form [2] for P(m) or [4] for Q(m) ? Why ? Why did Harris
want
to factor his polynomial in that form? If that form of
factorization
cannot be justified, then what Harris and the rest of us have
been
arguing about for the last several months is academic and we
all look
pretty silly. [Note this is likely true in any case!]
That factorizations [5] and [6] can be assumed is a
consequence
of a theorem of Magidin and McKinnon, though evidently the
result
was proved earlier by someone else. In general as far as I can
tell, this theorem does not imply factorizations [2] or [4].
Perhaps Arturo can shed some light on this.
>
>
> And where exactly is your proof that this cannot happen?
> You are rather stupid when it comes to math you disgusting
piece of
> trash.
> You canÕt have xy=26 in the ring of algebraic integers 
with
x varying
> over algebraic integers *and* y an algebraic integer for
all x.
In the setting we are discussing, x is w1(m) and 26 is
f^2. You are assuming that w1(m), as a function of m, takes
on an unlimited range of values as m ranges from 0 through all
positive integers. There is no reason to assume this. The
range
of w1(m), [for fixed values of f and u] may be quite limited.
If my statement of yesterday were true, w1(m) = f^{2/3} for
most values of m. Thus for fixed u and f, w1(m) would be 
almost
a constant function.
> YouÕre stupid, and despicable.
> YouÕre trash Nora Baron.
This is a new low even for you. I assume you have been
drinking. Try getting back to the math.
Nora B.
> James Harris
===
Subject: Re: Core error argument objection refuted, short
...
> Harris starts with the polynomial
> [1] P(m) = f^2 * [(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
> - 3*(-1 + m*f^2)*u^2*x + u*3*f]
...
> [2] P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
...
> Here is what I was thinking. Let Q(m) = P(m) / f^2, i.e.,
> [3] Q(m) = (m^3*f^4 -3*m^2*f^2 + 3*m)*x^3 - 3*(-1 +
m*f^2)*u^2*x +
u*3*f]
> Paralleling Harris, I assumed that Q(m) could be factored
in the form
> [4] Q(m) = (b1*x + u*f^{1/3})*(b2*x + u*f{1/3})*(b3*x +
u*f^{1/3}).
LetÕs see. First off, I think you are only interested in the
case where
the polynomial for the aÕs is irreducible. Because if it is
reducible
in linear factors, indeed exactly two aÕs are divisible by 
f.
If it is
reducible to the product of a linear term and a quadratic
term, one
of the aÕs is either a multiple of f or co-prime to f. In
both cases
the other two aÕs both are not coprime to f, and there is a
case shown
(with a1 divisible by f) where the factor the other two share
with f
is *not* something like f^{1/2}. It is in the example given:
(-sqrt(7) +- sqrt(15))/2.
(I do not know whether it is possible that the a from the
linear term is
coprime to f, but I think that when it is possible, something
similar
applies.)
Well now, assume irreducable. LetÕs build some 
defining
polynomial for
the bÕs. Assumming b1 to b3 are all algebraic integer would
lead to
the conclusion that (x-b1)(x-b2)(x-b3) is monic with
algebraic integer
coefficients. However, we do not know that already. We know
that when
we multiply each factor by f^{2/3} (or the complete
polynomial by f^2)
that the result has algebraic integer coefficients. So we get
(replacing
b1*f^{2/3} by a1, etc.):
f^2.x^3 - (a1+a2+a3).f^{4/3}.x^2 +
(a1.a2+a2.a3+a1.a3).f^{2/3}.x +
a1.a2.a3.
However, because we know the cubic of which the aÕs are 
roots:
x^3 + 3(-1+mf^2)x^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)
we know that of the polynomial for the bÕs the 
first, third
and fourth
term can be divided by f^2 (the third term is 0, the fourth
term is a
multiple of f^2). Furthermore we know what (a1+a2+a3) is:
3(1-m.f^2).
So if the assumption is correct (the bÕs defined 
above are
algebraic
integers),
3(1-m.f^2).f^{4/3}
must be divisible by f^2. It is however easily seen that this
is in
general
not the case. 3.f^{4/3}/f^2 = 3.f{-2/3} is not an algebraic
integer in
general. So the three aÕs are not divisible by f^{2/3} in
general.
> Possibly in that case also the
> most we can assume is
> [6] P(m) = (a1*x + d1)*(a2*x + d2)*(a3*x + d3),
> where a1, a2, a3 and d1, d2, d3 are algebraic integers.
That is,
> again, how can Harris assume that each of the dÕs is equal
to u*f ?
That assumption is a theorem, proven (quite a long time ago)
by Arturo
(it is generally called the Magidin-McKinnon theorem).
My assumption (and I think also ArturoÕs assumption) has 
been
that
we can write f = p.q.r where p, q and r are algebraic
integers,
a1 divisible by p.q, a2 divisible by p.r and a3 divisible by
q.r,
in the non-degenerate cases. I think this also covers the
degenerate
cases. E.g. if there is a linear decomposition we can have p
= f,
q = 1 and r = 1 and find that exactly two of the 
aÕs are
divisible
by f!
But it is much more difficult to prove *this*. The reason is
that
the decomposition of f in p, q and r needed depends on the
value of
m.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Core error argument objection refuted, short
Adjunct Assistant Professor at the University of Montana.
[.snip.]
>My assumption (and I think also ArturoÕs assumption) has
been that
>we can write f = p.q.r where p, q and r are algebraic
integers,
>a1 divisible by p.q, a2 divisible by p.r and a3 divisible by
q.r,
>in the non-degenerate cases. I think this also covers the
degenerate
>cases. E.g. if there is a linear decomposition we can have p
= f,
>q = 1 and r = 1 and find that exactly two of the 
aÕs are
divisible
>by f!
>But it is much more difficult to prove *this*. The reason is
that
>the decomposition of f in p, q and r needed depends on the
value of
I was going through it in an e-mail reply to some questions
Nora
sent. You can get it from the prime factorization of f and
some Galois
Theory applied to the splitting field, at least when the
polynomial is
irreducible.
What follows is almost certainly harder than it needs to
be... ThatÕs
my usual m.o.
Assume the polynomial (v^3+1)X^3 - 3vX + 1 is irreducible
over Q. Let
K be its splitting field, and let A be the ring of integers of
K. Then
[K:Q]=3 or [K:Q]=6 (in JamesÕs old situation, one could 
prove
that it
was always 6, because the discriminant was negative; this new
polynomial and the fewer restrictions on m and f do not seem
to let me
reach the same conclusion).
We have a1, a2, a3 in A, with
(x-a1)(x-a2)(x-a3) = x^3 + 3vx^2 - (v^3+1)
Note that
a1*a2*a3 = v^3+1
a1*a2 + a1*a3 + a2*a3 = 0
a1+a2+a3 = -3v.
where v = -1+mf^2
Note that if x divides f and divides all a1, a2, a3, then x
is a unit:
because -3v would be a multiple of x, and is coprime to f.
Note that if x=gcd(a1,f), moding out the second equation by x
we get
a2*a3 = 0 (mod x), so x cannot be coprime to both a2 and a3.
Analogous
arguments hold for gcd(a2,f) and gcd(a3,f).
Now factor (f) into prime ideals. Since K is Galois, the
splitting
must be a product of r different ideals, all raised to the
same
exponent e, and with the property that r*e divides [K:Q]. So
there are only
the following possibilities:
[1] (f) = p1*p2*p3*p4*p5*p6 (only if [K:Q]=6
[2] (f) = p1*p2*p3
[3] (f) = p1*p2 (only if [K:Q] = 6)
[4] (f) = p1
[5] (f) = p1^2*p2^2*p3^2 (only if [K:Q]=6)
[6] (f) = p1^2 (only if [K:Q]=6)
[7] (f) = p1^3*p2^3 (only if [K:Q]=6)
[8] (f) = p1^3
[9] (f) = p1^6 (only if [K:Q]=6).
Now consider the prime factorizations of (a1), (a2), and
(a3). They
will all be a product of some of the primes dividing (f) times
something coprime to (f). Let us ignore the primes coprime to
f for
simplicity.
Since the Galois group is transitivie on the primes and on
{a1, a2,
a3}, the kind of primes that show up must be the same. That
is, if
one prime is raised to the second power in (a1), then each
prime must
be raised to the second power in at least one of (a1), (a2),
(a3). And the number of primes in each will be the same, etc.
Since (a1)(a2)(a3) is f^2 times something coprime to f, the
number of
primes that show up in (f)^2 must be divisible by 3, so they
can be
split off evenly among all of (a1), (a2), (a3).
That means that we cannot be in situations [3], [4], or [6].
Since we
cannot have a prime dividing all of (a1), (a2), and (a3), we
cannot be
in situation [8] or [9] either.
If we were in situation [7], each of (ai) gets 4 primes,
which means
all of (a1), (a2), (a3) are divisible by p1*p2, which is also
impossible.
That leaves only situations [1], [2], or [5].
[1] (f) = p1*p2*p3*p4*p5*p6 (only if [K:Q]=6
[2] (f) = p1*p2*p3
[5] (f) = p1^2*p2^2*p3^2 (only if [K:Q]=6)
In situation [1], each of (ai) is divisible by 4 primes from
among p1,
p1, p2, p2, p3, p3, p4, p4, p5, p5, p6, p6. If (a1) were
divisible by
the square of p1, say, then the square of p2 would have to
divide one
of (a1), (a2), (a3), same for the square of each of the other
primes,
which would mean that (a1), (a2), and (a3) are pairwise
coprime, which
is impossible as seen above. So we must have something like
(a1) = p1*p2*p3*p4
(a2) = p1*p2*p5*p6
(a3) = p3*p4*p5*p6
as claimed.
In situation [5], no (ai) can be divisible by the 4th power
of one of
p1, p2, or p3, because then we would have (ai) = pi^4,
pairwise
coprime. So either we have (a1) = p1^2*p2^2, (a2)=p1^2*p3^2,
(a3)=p2^2*p3^2, as desired, or else we would have something
like
(a1)=p1^3*p2, (a2)=p2^3*p3, (a3)=p3^3*p1.
But since the extension must be of degree 6, the action of
Gal(K/Q) on
{a1,a2,a3} is doubly transitive, the autmorphism that fixes a1
and
exchanges a2 and a3 must fix p1 and fix p2 (since 
it fixes
(a1)), but
must also send p2 to p3 and p3 to p1 (since it sends (a2) to
(a3));
this is impossible, so we must have gcd(a1,f)=gcd(a1,f^2) and
again we
are in the nice situation.
Finally, in situation [3] each of (a1), (a2), (a3) must be
divisible
by 2 primes. It cannot be that, say, (a1)=p1^2, because then
they
would be pairwise coprime, which is impossible, so we must
have
(a1)=p1*p2, (a2)=p1*p3, (a3)=p2*p3 (up to relabeling the
primes),
which again yields the decomposition claimed.
When the polynomial is reducible, we have two possibilities:
either it
is a product of three linears, or it is a product of a linear
and an
irreducible quadratic.
If it is a product of three linear terms, you cannot have
that a1 is a
multiple of f^2 and a2 and a3 are coprime to f, since
a1*a2+a1*a3+a2*a3=0; and you cannot have all three multiples
of f, so
you get what James claims: if it splits completely, we get
two of a1,
a2, a3 multiples of f, one coprime to f.
If it splits into a product of a linear and an irreducible
quadratic,
then if f does not divide the linear root, then f^2 gets
split off
between the other two and it is coprime to the linear root;
then we
have f=r1*r2, r1 and r2 coprime. We cannot have a1 a multiple
or r1^2
and a2 a multiple of r2^2, so we must have both a1 and a2
multiples of
f, a3 coprime to f. In this case, we are writing f=f*1*1, and
get the
decomposition above.
If it splits into a linear and an irreducible quadratic, and
f divides
the linear root, then we cannot have f^2 dividing the linear
term
(because then you would have the other two coprime to f,
contradicting
that a1*a2+a1*a3+a2*a3=0), so f divides the linear root, and
the other
f is split among a1 and a2. So we have either a1,a2 multiples
of
sqrt(f), a3 of f; but then sqrt(f) divides all three which is
impossible; or else f=r1*r2 with r1 and r2 coprime, a1 a
multiple of
r1, a2 a multiple of r2, and a3 a multiple of f. (This is
what happens
with m=1, f=2). Then we are writing f=r1*r2*1, and we get the
decomposition given above.
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A manÕs capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A
great
many people are staggered to this extent, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Core error argument objection refuted, short
...
> You lack character and are a base liar.
> YouÕre a despicable human being.
> YouÕre disgusting trash.
Ah, you are describing yourself methinks.
> Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m),
so you
have
>
> xy=26
>
> And where exactly is your proof that this cannot happen?
> You are rather stupid when it comes to math you disgusting
piece of
> trash.
> You canÕt have xy=26 in the ring of algebraic integers 
with
x varying
> over algebraic integers *and* y an algebraic integer for
all x.
No, nobody has ever said that. x is a function of m,
remember? How do
you conclude that x ranges over *all* algebraic integers?
> YouÕre stupid, and despicable.
> YouÕre trash Nora Baron.
A description of JSH?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Core error argument objection refuted, short
> ...
> > You lack character and are a base liar.
> > YouÕre a despicable human being.
> > YouÕre disgusting trash.
> Ah, you are describing yourself methinks.
> > Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m),
so you
have
> > xy=26
> > And where exactly is your proof that this cannot happen?
> > You are rather stupid when it comes to math you
disgusting piece of
> > trash.
> > You canÕt have xy=26 in the ring of algebraic integers
with x varying
> > over algebraic integers *and* y an algebraic integer for
all x.
> No, nobody has ever said that. x is a function of m,
remember? How do
> you conclude that x ranges over *all* algebraic integers?
> > YouÕre stupid, and despicable.
Sounds like something my 7 year old brother would say, not a
man in his
30s.
> > YouÕre trash Nora Baron.
> A description of JSH?
> --
> dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
> home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Core error argument objection refuted, short
<deleted Also incorrect and false. Remember what you said
above:
... consider w_1(m), a factor of a_1 that is a factor of f...
If w_1(m) is a factor of f, that can only mean f/w_1(m) is
> an algebraic integer, which of course implies that uf/w_1(m)
> is an algebraic integer: this is your *assumption* here.
> The assumption is itself a contradiction, in that it
requires the
> ability to define one algebraic integer function by the
multiplicative
> inverse of another algebraic integer function.
> Basic algebra.
> IF THIS IS BASIC ALGEBRA, WHY CANÕT I UNDERSTAND YOUR
ARGUMENT? I
UNDERSTAND
> ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.
Unfortunately the people who *do* understand it are doing
their best
to confuse everyone else, and youÕve done your part by
basically
heckling me with posts.
If I didnÕt have mathematicians fighting me, 
then I could
carefully go
through each and every detail to work it out so that everyone
who has
a basic grasp of algebra can understand.
But first I need you to help me, by NOT letting posters like
Nora
Baron and Arturo Magidin get away with muddying the waters.
Other posters need to begin challenging these people who have
betrayed
you, after all, theyÕve had many of you nodding along with
something
as dumb as xy=2 with *both* x and y algebraic integers where
x varies
over the algebraic integers which you get using u=1, f=2, and
the
assumption of a w_1(m) as a factor of f thatÕs a varying
factor of
a_1, with x=w_1(m), y=uf/w_1(m).
ItÕs posters like Arturo Magidin and Nora Baron 
whoÕve helped
make
mathematicians look like fools and patsies, willing to lie and
challenge mathematics itself either willingly, or from
incompetence.
James Harris
===
Subject: Re: Core error argument objection refuted, short
> <deleted Also incorrect and false. Remember what you said
above:
... consider w_1(m), a factor of a_1 that is a factor of f...
If w_1(m) is a factor of f, that can only mean f/w_1(m) is
> an algebraic integer, which of course implies that uf/w_1(m)
> is an algebraic integer: this is your *assumption* here.
The assumption is itself a contradiction, in that it requires
the
> ability to define one algebraic integer function by the
multiplicative
> inverse of another algebraic integer function.
Basic algebra.
> IF THIS IS BASIC ALGEBRA, WHY CANÕT I UNDERSTAND YOUR
ARGUMENT? I
UNDERSTAND
> ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.
> Unfortunately the people who *do* understand it are doing
their best
> to confuse everyone else, and youÕve done your part by
basically
> heckling me with posts.
> If I didnÕt have mathematicians fighting me, 
then I could
carefully go
> through each and every detail to work it out so that
everyone who has
> a basic grasp of algebra can understand.
> But first I need you to help me, by NOT letting posters like
Nora
> Baron and Arturo Magidin get away with muddying the waters.
> Other posters need to begin challenging these people who
have betrayed
> you, after all, theyÕve had many of you nodding along with
something
> as dumb as xy=2 with *both* x and y algebraic integers
where x varies
> over the algebraic integers which you get using u=1, f=2,
and the
> assumption of a w_1(m) as a factor of f thatÕs a varying
factor of
> a_1, with x=w_1(m), y=uf/w_1(m).
> ItÕs posters like Arturo Magidin and Nora Baron 
whoÕve
helped make
> mathematicians look like fools and patsies, willing to lie
and
> challenge mathematics itself either willingly, or from
incompetence.
> James Harris
ItÕs you who are subjecting yourself to the heckling. If you
want the
heckling to stop, either stop posting, or stop the personal
attacks. I can
think of a much better way to do what youÕre doing, but will
you listen to
me? Probably not.
David Moran
===
Subject: Re: Core error argument objection refuted, short
> <deleted Also incorrect and false. Remember what you said
above:
... consider w_1(m), a factor of a_1 that is a factor of
f...
If w_1(m) is a factor of f, that can only mean f/w_1(m) is
> an algebraic integer, which of course implies that uf/w_1(m)
> is an algebraic integer: this is your *assumption* here.
The assumption is itself a contradiction, in that it requires
the
> ability to define one algebraic integer function by the
multiplicative
> inverse of another algebraic integer function.
Basic algebra.
> IF THIS IS BASIC ALGEBRA, WHY CANÕT I UNDERSTAND YOUR
ARGUMENT? I
> UNDERSTAND
> ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.
> Unfortunately the people who *do* understand it are doing
their best
> to confuse everyone else, and youÕve done your part by
basically
> heckling me with posts.
> If I didnÕt have mathematicians fighting me, 
then I could
carefully go
> through each and every detail to work it out so that
everyone who has
> a basic grasp of algebra can understand.
> But first I need you to help me, by NOT letting posters like
Nora
> Baron and Arturo Magidin get away with muddying the waters.
> Other posters need to begin challenging these people who
have betrayed
> you, after all, theyÕve had many of you nodding along with
something
> as dumb as xy=2 with *both* x and y algebraic integers
where x varies
> over the algebraic integers which you get using u=1, f=2,
and the
> assumption of a w_1(m) as a factor of f thatÕs a varying
factor of
> a_1, with x=w_1(m), y=uf/w_1(m).
> ItÕs posters like Arturo Magidin and Nora Baron 
whoÕve
helped make
> mathematicians look like fools and patsies, willing to lie
and
> challenge mathematics itself either willingly, or from
incompetence.
> James Harris
> ItÕs you who are subjecting yourself to the heckling. If
you want the
> heckling to stop, either stop posting, or stop the personal
attacks. I
can
> think of a much better way to do what youÕre doing, but
will you listen
to
> me? Probably not.
> David Moran
Readers hereÕs what I was replying to from this *same*
poster, which
IÕm copying down from this post.
> IF THIS IS BASIC ALGEBRA, WHY CANÕT I UNDERSTAND YOUR
ARGUMENT? I
> UNDERSTAND
> ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.
So I reply thinking that maybe the guy is serious, and you
can see
what he said in return, and notice his arrogance and lack of
social
graces.
ItÕs like IÕm dealing with children with VERY 
bad manners.
James Harris
===
Subject: Re: Core error argument objection refuted, short
<deleted
> Also incorrect and false. Remember what you said above:
... consider w_1(m), a factor of a_1 that is a factor of
f...
If w_1(m) is a factor of f, that can only mean f/w_1(m)
is
> an algebraic integer, which of course implies that uf/w_1(m)
> is an algebraic integer: this is your *assumption* here.
The assumption is itself a contradiction, in that it requires
the
> ability to define one algebraic integer function by the
multiplicative
> inverse of another algebraic integer function.
Basic algebra.
> IF THIS IS BASIC ALGEBRA, WHY CANÕT I UNDERSTAND YOUR
ARGUMENT? I
> UNDERSTAND
> ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.
Unfortunately the people who *do* understand it are doing
their best
> to confuse everyone else, and youÕve done your part by
basically
> heckling me with posts.
If I didnÕt have mathematicians fighting me, 
then I could
carefully
go
> through each and every detail to work it out so that
everyone who has
> a basic grasp of algebra can understand.
But first I need you to help me, by NOT letting posters like
Nora
> Baron and Arturo Magidin get away with muddying the waters.
Other posters need to begin challenging these people who have
betrayed
> you, after all, theyÕve had many of you nodding along with
something
> as dumb as xy=2 with *both* x and y algebraic integers
where x varies
> over the algebraic integers which you get using u=1, f=2,
and the
> assumption of a w_1(m) as a factor of f thatÕs a varying
factor of
> a_1, with x=w_1(m), y=uf/w_1(m).
ItÕs posters like Arturo Magidin and Nora Baron 
whoÕve helped
make
> mathematicians look like fools and patsies, willing to lie
and
> challenge mathematics itself either willingly, or from
incompetence.
> James Harris
> ItÕs you who are subjecting yourself to the heckling. If
you want the
> heckling to stop, either stop posting, or stop the personal
attacks. I
can
> think of a much better way to do what youÕre doing, but
will you listen
to
> me? Probably not.
> David Moran
> Readers hereÕs what I was replying to from this *same*
poster, which
> IÕm copying down from this post.
> IF THIS IS BASIC ALGEBRA, WHY CANÕT I UNDERSTAND YOUR
ARGUMENT? I
> UNDERSTAND
> ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.
> So I reply thinking that maybe the guy is serious, and you
can see
> what he said in return, and notice his arrogance and lack
of social
> graces.
> ItÕs like IÕm dealing with children with 
VERY bad manners.
> James Harris
Hey, IÕm not arrogant to anyone who doesnÕt 
deserve it. You
need to grow
up,
James. Your name calling makes me think IÕm reading 
something
from my 7
year
old brother. I can be nice, or I can be mean; you decide.
David Moran
===
Subject: Simple principle, core error proven
Luckily for me the mathematical argument that proves that
IÕve been
correct all along can be further simplified by the use of
*numbers*
instead of variables, as while algebra as an idea is
well-founded, so
letter symbols should do, when mathematicians are lying about
the
mathematics, you need to use what you can, and pray.
1. First the problematic definition:
Algebraic integers are defined to be roots of monic
polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17,
where
monic refers to the leading coefficient.
My assertion is that the over hundred year old definition
excludes
numbers that have to be included to keep from having
contradiction
i.e. mathematical inconsistency.
2. The important tool I use is a polynomial:
P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7)
where m varies in the ring of algebraic integers.
Some may find it looks odd. But the entire point of that form
is to
do something a little different than anyone else apparently
has done
before, which is to get non-polynomial factors.
And the factorization with those factors is
P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)
where the aÕs are roots of the following cubic:
a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m).
So you can see that the aÕs are functions of m, and if you
really want
headaches you can go ahead and solve the cubic to get a view
of them.
However, I can move on without needing their explicit form.
g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7)
and I can find those terms that are free of m by setting m=0,
just
like with any polynomial, you know like with S(m) = 2(m^2 +
2m + 1),
S(0) = 2, which you can just look at and see here, but with
my P(m)
itÕs just a little harder, so I set m=0, which gives me
P(0) = 49(3(5) + 7),
which fits with the cubic as at m=0 it gives
a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3,
to show that at m=0, the three factors are
g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22.
Now dividing P(m) by 49 gives
P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7
and the question is what happens to the gÕs, but look now at
P(0)/49
as that is
P(0)/49 = 3(5) + 7
as two factors of 7, each 7, have beeen divided off, which is
easy to
see.
The only way that can happen is if
P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
where only two of the aÕs have 7 as a factor, where the idea
is almost
trivially simple as consider a polynomial like
S(m) = 7(m^2 + 2m + 1) = (b_1 m + 7)(b_2 m + 1)
and notice that S(0) = 7, while S(m)/7, gives you
S(m)/7 = m^2 + 2m + 1,
which means that
S(m)/7 = (b_1 m/7 + 1)(b_2 m + 1) = m^2 + 2m + 1.
ItÕs the same basic idea while what I have is more
complicated as IÕm
showing you an over hundred year old ßaw, and it turns out
that it
takes a somewhat complicated expression to show it which
P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7
is, but the same principle works as just like with S(m), with
P(m),
dividing out 49, affects the independent or constant terms,
revealing
factors of the aÕs.
That is, from the distributive property, factors that are 7
must
divide through.
So now I know that the correct factorization is
P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
which is like
S(m)/7 = (b_1 m/7 + 1)(b_2 m + 1).
3. So *two* of the aÕs *should* have 7 as a factor, and in
fact they
do, in a proper ring, but the ring of algebraic integers has
problems,
so that for certain values of m, they wonÕt.
ItÕs that *inconsistency* which shows you 
thereÕs a problem
because
mathematics isnÕt about being wishy-washy, where sometimes
something
works and then other times it doesnÕt.
That error has sat in mathematics--the body of discoveries
commonly
called mathematics--for over a *hundred* years.
Since I found the error I should probably get rich and famous
from it,
but so far mathematicians IÕve contacted seem more 
interested
in
denying or hiding the error than in telling the truth.
However, that
means there is this error, which may sink lots of proofs over
the
past hundred years in an area of mathematics called algebraic
number
theory.
And in fact, mathematicians may be engaging in a bigger fraud
because
they may *know* by now just how big the problem is, and it
may be a
bigger embarrassment than even I am aware of at this point.
They may see themselves as having no choice but to hide it to
keep
their money, their prestige, and their history as it is known
to most
as of now.
Some of them may be trying to hide it partly out of envy or
jealousy
of my discovery as well. Mathematicians can be VERY vicious
for petty
and childish reasons IÕve found.
If you are a math student, you probably will want to stay out
of the
area of algebraic number theory, or consider carefully things
your
professors supposedly prove in the area.
While mathematicians behave this way, you have to wonder now
about
what they teach you.
James Harris
-------------------------
Intellectual laziness is about deciding
ahead of time what you wish to believe,
and daring God to be different.
http://lostincomment.blogspot.com/
===
Subject: Re: Simple principle, core error proven
Adjunct Assistant Professor at the University of Montana.
>Luckily for me the mathematical argument that proves that
IÕve been
>correct all along can be further simplified by the use of
*numbers*
>instead of variables, as while algebra as an idea is
well-founded, so
>letter symbols should do, when mathematicians are lying
about the
>mathematics, you need to use what you can, and pray.
>1. First the problematic definition:
>Algebraic integers are defined to be roots of monic
polynomials with
>integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17,
where
>monic refers to the leading coefficient.
>My assertion is that the over hundred year old definition
excludes
>numbers that have to be included to keep from having
contradiction
>i.e. mathematical inconsistency.
>2. The important tool I use is a polynomial:
>P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7)
>where m varies in the ring of algebraic integers.
>Some may find it looks odd. But the entire point of that form
is to
>do something a little different than anyone else apparently
has done
>before, which is to get non-polynomial factors.
>And the factorization with those factors is
>P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)
>where the aÕs are roots of the following cubic:
>a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m).
>So you can see that the aÕs are functions of m, and if you
really want
>headaches you can go ahead and solve the cubic to get a view
of them.
>However, I can move on without needing their explicit form.
>g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7)
>and I can find those terms that are free of m by setting m=0,
just
>like with any polynomial, you know like with S(m) = 2(m^2 +
2m + 1),
>S(0) = 2, which you can just look at and see here, but with
my P(m)
>itÕs just a little harder, so I set m=0, which gives me
>P(0) = 49(3(5) + 7),
>which fits with the cubic as at m=0 it gives
>a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3,
>to show that at m=0, the three factors are
>g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22.
>Now dividing P(m) by 49 gives
>P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7
>and the question is what happens to the gÕs, but look now 
at
P(0)/49
>as that is
>P(0)/49 = 3(5) + 7
>as two factors of 7, each 7, have beeen divided off, which
is easy to
>see.
At m=0.
>The only way that can happen is if
>P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)
And this is the false step. This is not the only way that can
happen
for arbitrary m, no matter how much you claim it is the only
way it
happen.
[.rest deleted.]
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A manÕs capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A
great
many people are staggered to this extent, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu